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# Solved Examples on Trigonometric Equations

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve the trigonometric equations
Determine the specific solutions of the trigonometric equations in the specified intervals
Write the general solution of the trigonometric equations
State the reasons for every step
Show all work
Unless otherwise specified or implied, only the exact values are allowed.

(1.) ACT For all real values of $x$, which of the following equations is true?

$F.\:\: \sin(7x) + \cos(7x) = 7 \\[3ex] G.\:\: \sin(7x) + \cos(7x) = 1 \\[3ex] H.\:\: 7\sin(7x) + 7\cos(7x) = 14 \\[3ex] J.\:\: \sin^2(7x) + \cos^2(7x) = 7 \\[3ex] K.\:\: \sin^2(7x) + \cos^2(7x) = 1$

This question should take you at most $5$ seconds.

$F.\:\: \sin(7x) + \cos(7x) = 7... NOT \\[3ex] G.\:\: \sin(7x) + \cos(7x) = 1...NOT \\[3ex] H.\:\: 7\sin(7x) + 7\cos(7x) = 14...NOT \\[3ex] J.\:\: \sin^2(7x) + \cos^2(7x) = 7...NOT \\[3ex] K.\:\: \sin^2(7x) + \cos^2(7x) = 1...Pythagorean\:\: Identity...YES$
(2.) Determine all the solutions of $4\sin \theta + 5 = 3$ in the interval $[0, 360^\circ)$

$4\sin \theta + 5 = 3 \\[3ex] 4\sin \theta = 3 - 5 \\[3ex] 4\sin \theta = -2 \\[3ex] \sin \theta = -\dfrac{2}{4} \\[5ex] \sin \theta = -\dfrac{1}{2} \\[5ex] Argument = \theta \\[3ex] \theta = \sin^{-1} \left(-\dfrac{1}{2}\right) \\[5ex] \theta = 210^\circ, 330^\circ ...\:Unit\:\: Circle\:\: Trig \\[3ex] 210 = 210 * \dfrac{\pi}{180} = \dfrac{7\pi}{6} \\[5ex] 330 = 330 * \dfrac{\pi}{180} = \dfrac{11\pi}{6} \\[5ex] \theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6} \\[5ex] \underline{General\:\: solutions} \\[3ex] \theta = 210 + 360k ... coterminal\:\: \angle s \\[3ex] \theta = 330 + 360k ... coterminal\:\: \angle s \\[3ex]$ Check
 $\underline{LHS} \\[3ex] \theta = 210^\circ \\[3ex] \\[3ex] 4\sin \theta + 5 \\[3ex] = 4 * \sin210 + 5 \\[5ex] = 4 * -\dfrac{1}{2} + 5 \\[5ex] = -2 + 5 \\[3ex] = 3 \\[3ex]$ $\theta = 330^\circ \\[3ex] 4\sin \theta + 5 \\[3ex] = 4 * \sin 330 + 5 \\[3ex] = 4 * -\dfrac{1}{2} + 5 \\[5ex] = -2 + 5 \\[3ex] = 3$ $\underline{RHS} \\[3ex] 3$
(3.) Determine all the solutions of $2\sin\beta - \sqrt{3} = 0$ in the interval $[0, 2\pi)$

$2\sin\beta - \sqrt{3} = 0 \\[3ex] 2\sin\beta = \sqrt{3} \\[3ex] \sin\beta = \dfrac{\sqrt{3}}{2} \\[5ex] Argument = \beta \\[3ex] \beta = \sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right) \\[5ex] \beta = 60^\circ, 120^\circ ...Unit\:\: Circle\:\: Trig \\[3ex] 60 = 60 * \dfrac{\pi}{180} = \dfrac{\pi}{3} \\[5ex] 120 = 120 * \dfrac{\pi}{180} = \dfrac{2\pi}{3} \\[5ex] \underline{General\:\: solutions} \\[3ex] \beta = \dfrac{\pi}{3} + 2\pi k ... coterminal\:\: \angle s \\[5ex] \beta = \dfrac{2\pi}{3} + 2\pi k ... coterminal\:\: \angle s \\[5ex]$ Check
 $\underline{LHS} \\[3ex] \beta = \dfrac{\pi}{3} \\[5ex] 2\sin\beta - \sqrt{3} \\[3ex] = 2 * \sin\dfrac{\pi}{3} - \sqrt{3} \\[5ex] = 2 * \dfrac{\sqrt{3}}{2} - \sqrt{3} \\[5ex] = \sqrt{3} - \sqrt{3} \\[3ex] = 0 \\[3ex]$ $\beta = \dfrac{2\pi}{3} \\[5ex] 2\sin\beta - \sqrt{3} \\[3ex] = 2 * \sin\dfrac{2\pi}{3} - \sqrt{3} \\[5ex] = 2 * \dfrac{\sqrt{3}}{2} - \sqrt{3} \\[5ex] = \sqrt{3} - \sqrt{3} \\[3ex] = 0$ $\underline{RHS} \\[3ex] 0$
(4.) ACT What are the values of $\theta$, between $\theta$ and $2\pi$, when $\tan \theta = -1$?

$A.\:\: \dfrac{\pi}{4} \:\:and\:\: \dfrac{3\pi}{4} \:\:only \\[5ex] B.\:\: \dfrac{3\pi}{4} \:\:and\:\: \dfrac{5\pi}{4} \:\:only \\[5ex] C.\:\: \dfrac{3\pi}{4} \:\:and\:\: \dfrac{7\pi}{4} \:\:only \\[5ex] D.\:\: \dfrac{5\pi}{4} \:\:and\:\: \dfrac{7\pi}{4} \:\:only \\[5ex] E.\:\: \dfrac{\pi}{4}, \:\dfrac{3\pi}{4}, \:\dfrac{5\pi}{4}, \:\:and\:\: \dfrac{7\pi}{4}$

$\tan \theta = -1 \\[3ex] Argument = \theta \\[3ex] Assume\:\: \tan\theta = 1 \\[3ex] \theta = \tan^{-1}(1) = 45^\circ = 45 * \dfrac{\pi}{180} = \dfrac{\pi}{4} \\[5ex] But\:\: \tan\theta = -1 \\[3ex] \tan \:\:is\:\: negative\:\: in\:\: 2nd \:\:and\:\: 4th \:\:quadrants \\[3ex] -\tan \left(\dfrac{\pi}{4}\right) = \tan \left(\pi - \dfrac{\pi}{4}\right)...2nd\:\: Quadrant\:\: Identity \\[5ex] = \tan \left(\dfrac{4\pi}{4} - \dfrac{\pi}{4}\right) \\[5ex] = \tan \left(\dfrac{4\pi - \pi}{4}\right) \\[5ex] = \tan \left(\dfrac{3\pi}{4}\right) \\[5ex] Also, \\[3ex] -\tan \left(\dfrac{\pi}{4}\right) = \tan \left(2\pi - \dfrac{\pi}{4}\right)...4th\:\: Quadrant\:\: Identity \\[5ex] = \tan \left(\dfrac{8\pi}{4} - \dfrac{\pi}{4}\right) \\[5ex] = \tan \left(\dfrac{8\pi - \pi}{4}\right) \\[5ex] = \tan \left(\dfrac{7\pi}{4}\right) \\[5ex] \theta = \dfrac{3\pi}{4},\: \dfrac{7\pi}{4} \\[5ex]$ Check
 $\underline{LHS} \\[3ex] \tan \theta \\[3ex] \theta = \dfrac{3\pi}{4} \\[5ex] \tan \left(\dfrac{3\pi}{4}\right) = -1 \\[5ex]$ $\theta = \dfrac{7\pi}{4} \\[5ex] \tan \left(\dfrac{7\pi}{4}\right) = -1$ $\underline{RHS} \\[3ex] -1$
(5.) ACT If $0^\circ \le x^\circ \le 90^\circ$, and $2\sin^2x^\circ - 1 = 0$,
then $x^\circ =$?

$A.\:\: 0^\circ \\[3ex] B.\:\: 30^\circ \\[3ex] C.\:\: 45^\circ \\[3ex] D.\:\: 60^\circ \\[3ex] E.\:\: 90^\circ$

$2\sin^2x^\circ - 1 = 0 \\[3ex] 2\sin^2x^\circ = 1 \\[3ex] \sin^2x^\circ = \dfrac{1}{2} \\[3ex] \sin x = \pm \sqrt{\dfrac{1}{2}} \\[5ex] Argument = x \\[3ex] \sqrt{\dfrac{1}{2}} = \dfrac{\sqrt{1}}{\sqrt{2}} \\[5ex] = \dfrac{1}{\sqrt{2}} \\[5ex] = \dfrac{1}{\sqrt{2}} * \dfrac{\sqrt{2}}{\sqrt{2}} \\[5ex] = \dfrac{\sqrt{2}}{2} \\[5ex] \rightarrow \sin x = \pm \dfrac{\sqrt{2}}{2} \\[5ex] x = \sin^{-1}\left(-\dfrac{\sqrt{2}}{2}\right) \:\:OR\:\: x = \sin^{-1}\left(\dfrac{\sqrt{2}}{2}\right) \\[5ex] x = -45^\circ \:\:OR\:\: x = 45^\circ \\[3ex] Because\:\: 0^\circ \le x^\circ \le 90^\circ \\[3ex] x = 45^\circ \\[3ex]$ Check
 $\underline{LHS} \\[3ex] 2\sin^2x^\circ - 1 \\[3ex] x = 45^\circ \\[3ex] 2\sin^2(45) - 1 \\[3ex] 2 * (\sin 45)^2 - 1 \\[3ex] \sin 45 = \dfrac{\sqrt{2}}{2} \\[5ex] (\sin 45)^2 = \left(\dfrac{\sqrt{2}}{2}\right)^2 \\[5ex] (\sin 45)^2 = \dfrac{\sqrt{2}^2}{2^2} \\[5ex] (\sin 45)^2 = \dfrac{2}{4} = \dfrac{1}{2} \\[5ex] = 2 * \dfrac{1}{2} - 1 \\[5ex] = 1 - 1 \\[3ex] = 0$ $\underline{RHS} \\[3ex] 0$
(6.) ACT What are the values of $\theta$, between $0$ and $360^\circ$, when $\tan \theta = -1$?

$A.\:\: 225^\circ \:\:and\:\: 315^\circ \:\:only \\[3ex] B.\:\: 135^\circ \:\:and\:\: 315^\circ \:\:only \\[3ex] C.\:\: 135^\circ \:\:and\:\: 225^\circ \:\:only \\[3ex] D.\:\: 45^\circ \:\:and\:\: 135^\circ \:\:only \\[3ex] E.\:\: 45^\circ, \:135^\circ, \:225^\circ, \:\:and\:\: 315^\circ$

$\tan \theta = -1 \\[3ex] Argument = \theta \\[3ex] Assume\:\: \tan\theta = 1 \\[3ex] \theta = \tan^{-1}(1) = 45^\circ = 45 * \dfrac{\pi}{180} = \dfrac{\pi}{4} \\[5ex] But\:\: \tan\theta = -1 \\[3ex] \tan \:\:is\:\: negative\:\: in\:\: 2nd \:\:and\:\: 4th \:\:quadrants \\[3ex] -\tan 45 = \tan (180 - 45)...2nd\:\: Quadrant\:\: Identity \\[3ex] = \tan 135 \\[3ex] Also, \\[3ex] -\tan 45 = \tan (360 - 45)...4th\:\: Quadrant\:\: Identity \\[3ex] = \tan 315 \\[3ex] \theta = 135^\circ,\: 315^\circ \\[5ex]$ Check
 $\underline{LHS} \\[3ex] \tan \theta \\[3ex] \theta = 135^\circ \\[3ex] \tan 135 = -1 \\[3ex]$ $\theta = 315^\circ \\[3ex] \tan 315 = -1$ $\underline{RHS} \\[3ex] -1$
(7.) Determine all the solutions of $\sin\alpha - \cos\alpha = 0$ in the interval $[0, 2\pi)$

This is not straightforward. It has two functions - the sine function and the cosine function.
What do we do to express it in terms of only one function?

$\sin\alpha - \cos\alpha = 0 \\[3ex] Square\:\: both\:\: sides \\[3ex] (\sin\alpha - \cos\alpha)^2 = 0^2 \\[3ex] (\sin\alpha - \cos\alpha)(\sin\alpha - \cos\alpha) = 0 \\[3ex] \sin^2\alpha - \sin\alpha\cos\alpha - \sin\alpha\cos\alpha + \cos^2\alpha = 0 \\[3ex] \sin^2\alpha + \cos^2\alpha - 2\sin\alpha\cos\alpha = 0 \\[3ex] \sin^2\alpha + \cos^2\alpha = 1 ... Pythagorean\:\: Identity \\[3ex] 1 - 2\sin\alpha\cos\alpha = 0 \\[3ex] 1 = 2\sin\alpha\cos\alpha \\[3ex] 2\sin\alpha\cos\alpha = 1 \\[3ex] 2\sin\alpha\cos\alpha = \sin2\alpha ... Double\:\: Angle\:\: Formula \\[3ex] \sin2\alpha = 1 \\[3ex] Argument = 2\alpha \\[3ex] \therefore interval = [0, 4\pi) \\[3ex] 2\alpha = \sin^{-1}(1) \\[3ex] \sin^{-1}(1) = 90^\circ ... Unit\:\: Circle\:\: Trig \\[5ex] Also,\:\: based\:\: on\:\: [0, 4\pi) = [0, 4 * 180^\circ) = [0, 720^\circ) \\[3ex] \sin^{-1}(1) = 90 + 360(1) ...coterminal \angle s \\[3ex] \sin^{-1}(1) = 90 + 360 \\[3ex] \sin^{-1}(1) = 450^\circ \\[3ex] \therefore 2\alpha = 90, \:\:2\alpha = 450 \\[5ex] \alpha = \dfrac{90}{2}, \:\:\alpha = \dfrac{450}{2} \\[5ex] \alpha = 45^\circ, 225^\circ \\[3ex] 45^\circ = 45 * \dfrac{\pi}{180} = \dfrac{\pi}{4} \\[5ex] 225^\circ = 225 * \dfrac{\pi}{180} = \dfrac{5\pi}{4} \\[5ex] \alpha = \dfrac{\pi}{4}, \dfrac{5\pi}{4} \\[5ex]$ Check
 $\underline{LHS} \\[3ex] \sin\alpha - \cos\alpha \\[3ex] \alpha = \dfrac{\pi}{4} \\[5ex] \sin\alpha = \sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2} \\[5ex] \cos\alpha = \cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2} \\[5ex] = \dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{2}}{2} \\[5ex] = 0 \\[5ex]$ $\alpha = \dfrac{5\pi}{4} \\[5ex] \sin\alpha = \sin\dfrac{5\pi}{4} = -\dfrac{\sqrt{2}}{2} \\[5ex] \cos\alpha = \cos\dfrac{5\pi}{4} = -\dfrac{\sqrt{2}}{2} \\[5ex] = -\dfrac{\sqrt{2}}{2} - -\dfrac{\sqrt{2}}{2} \\[5ex] = -\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} \\[5ex] = 0$ $\underline{RHS} \\[3ex] 0$

$\underline{General\:\: solutions} \\[3ex] \alpha = \dfrac{\pi}{4} + 2\pi k ...coterminal\:\: \angle s \\[5ex] \alpha = \dfrac{5\pi}{4} + 2\pi k ...coterminal\:\: \angle s$
(8.) Determine all the solutions of $\cos\alpha + \sqrt{3}\sin\alpha = 1$ in the interval $[0, 2\pi]$

This is not straightforward. It has two functions - the cosine function and the sine function.
What do we do to express it in terms of only one function?

$\cos\alpha + \sqrt{3}\sin\alpha = 1 \\[3ex] \sqrt{3}\sin\alpha = 1 - \cos\alpha \\[3ex] Square\:\: both\:\: sides \\[3ex] (\sqrt{3}\sin\alpha)^2 = (1 - \cos\alpha)^2 \\[3ex] (\sqrt{3})^2 * \sin^2\alpha = (1 - \cos\alpha)(1 - \cos\alpha) \\[3ex] 3\sin^2\alpha = 1 - \cos\alpha - \cos\alpha + \cos^2\alpha \\[3ex] 3\sin^2\alpha = 1 - 2\cos\alpha + \cos^2\alpha \\[3ex] \sin^2\alpha + \cos^2\alpha = 1 ... Pythagorean\:\: Identity \\[3ex] \sin^2\alpha = 1 - \cos^2\alpha \\[3ex] 3(1 - \cos^2\alpha) = 1 - 2\cos\alpha + \cos^2\alpha \\[3ex] 3 - 3\cos^2\alpha = 1 - 2\cos\alpha + \cos^2\alpha \\[3ex] 0 = 1 - 2\cos\alpha + \cos^2\alpha - 3 + 3\cos^2\alpha \\[3ex] 1 - 2\cos\alpha + \cos^2\alpha - 3 + 3\cos^2\alpha = 0 \\[3ex] 4\cos^2\alpha - 2\cos\alpha - 2 = 0 \\[3ex] Divide\:\: all\:\: terms\:\: by\:\: 2 \\[3ex] 2\cos^2\alpha - \cos\alpha - 1 = 0 \\[3ex] Let\:\: \cos\alpha = p \\[3ex] 2p^2 - p - 1 = 0 \\[3ex] 2p^2 + p - 2p - 1 = 0 \\[3ex] p(2p + 1) - 1(2p + 1) = 0 \\[3ex] (2p + 1)(p - 1) = 0 \\[3ex] 2p + 1 = 0 \:\:\:OR\:\:\: p - 1 = 0 \\[3ex] 2p = -1 \:\:\:OR\:\:\: p = 1 \\[3ex] p = -\dfrac{1}{2} \:\:\:OR\:\:\: p = 1 \\[3ex] Substitute\:\: back \\[3ex] \rightarrow \cos\alpha = -\dfrac{1}{2} \\[3ex] \alpha = \cos^{-1}\left(-\dfrac{1}{2}\right) \\[5ex] \alpha = 120^\circ, 240^\circ ... Unit\:\: Circle\:\: Trig \\[3ex] \rightarrow \cos\alpha = 1 \\[3ex] \alpha = \cos^{-1}(1) \\[3ex] \alpha = 0^\circ, 360^\circ ... Unit\:\: Circle\:\: Trig \\[3ex]$ Check
 $\underline{LHS} \\[3ex] \cos\alpha + \sqrt{3}\sin\alpha \\[3ex] \alpha = 120 \\[5ex] \cos\alpha = \cos 120 = -\dfrac{1}{2} \\[5ex] \sin\alpha = \sin 120 = \dfrac{\sqrt{3}}{2} \\[5ex] \sqrt{3}\sin\alpha = \sqrt{3} * \dfrac{\sqrt{3}}{2} = \dfrac{3}{2} \\[5ex] = -\dfrac{1}{2} + \dfrac{3}{2} \\[5ex] = -\dfrac{1}{2} + \dfrac{3}{2} \\[5ex] = \dfrac{-1 + 3}{2} \\[5ex] = -\dfrac{2}{2} \\[5ex] = 1 \\[3ex] YES \\[3ex]$ $\alpha = 240 \\[5ex] \cos\alpha = \cos 240 = -\dfrac{1}{2} \\[5ex] \sin\alpha = \sin 240 = -\dfrac{\sqrt{3}}{2} \\[5ex] \sqrt{3}\sin\alpha = \sqrt{3} * -\dfrac{\sqrt{3}}{2} = -\dfrac{3}{2} \\[5ex] = -\dfrac{1}{2} + - \dfrac{3}{2} \\[5ex] = -\dfrac{1}{2} - \dfrac{3}{2} \\[5ex] = \dfrac{-1 - 3}{2} \\[5ex] = -\dfrac{4}{2} \\[5ex] = -2 \\[3ex] Extraneous\:\ root \\[3ex] NO \\[3ex]$ $\alpha = 0 \\[3ex] \cos\alpha = \cos 0 = 1 \\[3ex] \sin\alpha = \sin 0 = 0 \\[3ex] \sqrt{3}\sin\alpha = \sqrt{3} * 0 = 0\\[3ex] = 1 + 0 \\[5ex] = 1 \\[3ex] YES \\[3ex]$ $\alpha = 360 \\[3ex] \cos\alpha = \cos 360 = 1 \\[3ex] \sin\alpha = \sin 360 = 0 \\[3ex] \sqrt{3}\sin\alpha = \sqrt{3} * 0 = 0\\[3ex] = 1 + 0 \\[5ex] = 1 \\[3ex] YES$ $\underline{RHS} \\[3ex] 1$

$\underline{Specific\:\: solutions} \\[3ex] \alpha = 120^\circ = 120 * \dfrac{\pi}{180} = \dfrac{2\pi}{3} \\[5ex] \alpha = 0^\circ = 120 * \dfrac{\pi}{180} = 0 \\[5ex] \alpha = 360^\circ = 360 * 120 * \dfrac{\pi}{180} = 2\pi \\[5ex] \underline{General\:\: solutions} \\[3ex] \alpha = 0 + 2\pi k = 2\pi k ... coterminal\:\: \angle s \\[3ex] \alpha = \dfrac{2\pi}{3} + 2\pi k ... coterminal\:\: \angle s$
(9.) Determine all the solutions of $6\cos^2\theta = 3$ in the interval $[0, 2\pi)$

$6\cos^2\theta = 3 \\[3ex] \cos^2\theta = \dfrac{3}{6} = \dfrac{1}{2} \\[5ex] \cos\theta = \pm \sqrt{\dfrac{1}{2}} \\[5ex] Argument = \theta \\[3ex] \sqrt{\dfrac{1}{2}} = \dfrac{\sqrt{1}}{\sqrt{2}} \\[5ex] = \dfrac{1}{\sqrt{2}} \\[5ex] = \dfrac{1}{\sqrt{2}} * \dfrac{\sqrt{2}}{\sqrt{2}} \\[5ex] = \dfrac{\sqrt{2}}{2} \\[5ex] \rightarrow \cos \theta = \pm \dfrac{\sqrt{2}}{2} \\[5ex] \theta = \cos^{-1}\left(\dfrac{\sqrt{2}}{2}\right) \\[5ex] \theta = 45^\circ, 315^\circ ...Unit\:\: Circle\:\: Trig \\[3ex] 45 = 45 * \dfrac{\pi}{180} = \dfrac{\pi}{4} \\[5ex] 315 = 315 * \dfrac{\pi}{180} = \dfrac{7\pi}{4} \\[5ex] \theta = \cos^{-1}\left(-\dfrac{\sqrt{2}}{2}\right) \\[5ex] \theta = 135^\circ, 225^\circ ...Unit\:\: Circle\:\: Trig \\[3ex] 135 = 135 * \dfrac{\pi}{180} = \dfrac{3\pi}{4} \\[5ex] 225 = 225 * \dfrac{\pi}{180} = \dfrac{5\pi}{4} \\[5ex] \theta = \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{5\pi}{4}, \dfrac{7\pi}{4} \\[5ex] \underline{General\:\: solutions} \\[3ex] \beta = \dfrac{\pi}{4} + 2\pi k ... coterminal\:\: \angle s \\[5ex] \beta = \dfrac{3\pi}{4} + 2\pi k ... coterminal\:\: \angle s \\[5ex] \beta = \dfrac{5\pi}{4} + 2\pi k ... coterminal\:\: \angle s \\[5ex] \beta = \dfrac{7\pi}{4} + 2\pi k ... coterminal\:\: \angle s \\[5ex]$ Check
 $\underline{LHS} \\[3ex] 6\cos^2\theta \\[3ex] \theta = \dfrac{\pi}{4} \\[5ex] \cos\theta = \cos\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2} \\[5ex] \cos^2\theta = \left(\dfrac{\sqrt{2}}{2}\right)^2 = \dfrac{(\sqrt{2})^2}{2^2} = \dfrac{2}{4} = \dfrac{1}{2} \\[5ex] = 6 * \dfrac{1}{2} \\[5ex] = 3 \\[3ex]$ $\theta = \dfrac{3\pi}{4} \\[5ex] \cos\theta = \cos\dfrac{3\pi}{4} = -\dfrac{\sqrt{2}}{2} \\[5ex] \cos^2\theta = \left(-\dfrac{\sqrt{2}}{2}\right)^2 = \dfrac{(-\sqrt{2})^2}{2^2} = \dfrac{2}{4} = \dfrac{1}{2} \\[5ex] = 6 * \dfrac{1}{2} \\[5ex] = 3 \\[3ex]$ $\theta = \dfrac{5\pi}{4} \\[5ex] \cos\theta = \cos\dfrac{5\pi}{4} = \dfrac{\sqrt{2}}{2} \\[5ex] \cos^2\theta = \left(\dfrac{\sqrt{2}}{2}\right)^2 = \dfrac{(\sqrt{2})^2}{2^2} = \dfrac{2}{4} = \dfrac{1}{2} \\[5ex] = 6 * \dfrac{1}{2} \\[5ex] = 3 \\[3ex]$ $\theta = \dfrac{7\pi}{4} \\[5ex] \cos\theta = \cos\dfrac{7\pi}{4} = -\dfrac{\sqrt{2}}{2} \\[5ex] \cos^2\theta = \left(-\dfrac{\sqrt{2}}{2}\right)^2 = \dfrac{(-\sqrt{2})^2}{2^2} = \dfrac{2}{4} = \dfrac{1}{2} \\[5ex] = 6 * \dfrac{1}{2} \\[5ex] = 3$ $\underline{RHS} \\[3ex] 3$
(10.) Determine all the solutions of $\cos\alpha - \sqrt{3}\sin\alpha = 1$ in the interval $[0, 2\pi]$

This is not straightforward. It has two functions - the cosine function and the sine function.
What do we do to express it in terms of only one function?

$\cos\alpha - \sqrt{3}\sin\alpha = 1 \\[3ex] -\sqrt{3}\sin\alpha = 1 - \cos\alpha \\[3ex] Square\:\: both\:\: sides \\[3ex] (-1 * \sqrt{3}\sin\alpha)^2 = (1 - \cos\alpha)^2 \\[3ex] (-1)^2 * (\sqrt{3})^2 * \sin^2\alpha = (1 - \cos\alpha)(1 - \cos\alpha) \\[3ex] 1 * 3 * \sin^2\alpha = 1 - \cos\alpha - \cos\alpha + \cos^2\alpha \\[3ex] 3\sin^2\alpha = 1 - 2\cos\alpha + \cos^2\alpha \\[3ex] \sin^2\alpha + \cos^2\alpha = 1 ... Pythagorean\:\: Identity \\[3ex] \sin^2\alpha = 1 - \cos^2\alpha \\[3ex] 3(1 - \cos^2\alpha) = 1 - 2\cos\alpha + \cos^2\alpha \\[3ex] 3 - 3\cos^2\alpha = 1 - 2\cos\alpha + \cos^2\alpha \\[3ex] 0 = 1 - 2\cos\alpha + \cos^2\alpha - 3 + 3\cos^2\alpha \\[3ex] 1 - 2\cos\alpha + \cos^2\alpha - 3 + 3\cos^2\alpha = 0 \\[3ex] 4\cos^2\alpha - 2\cos\alpha - 2 = 0 \\[3ex] Divide\:\: all\:\: terms\:\: by\:\: 2 \\[3ex] 2\cos^2\alpha - \cos\alpha - 1 = 0 \\[3ex] Let\:\: \cos\alpha = p \\[3ex] 2p^2 - p - 1 = 0 \\[3ex] 2p^2 + p - 2p - 1 = 0 \\[3ex] p(2p + 1) - 1(2p + 1) = 0 \\[3ex] (2p + 1)(p - 1) = 0 \\[3ex] 2p + 1 = 0 \:\:\:OR\:\:\: p - 1 = 0 \\[3ex] 2p = -1 \:\:\:OR\:\:\: p = 1 \\[3ex] p = -\dfrac{1}{2} \:\:\:OR\:\:\: p = 1 \\[3ex] Substitute\:\: back \\[3ex] \rightarrow \cos\alpha = -\dfrac{1}{2} \\[3ex] \alpha = \cos^{-1}\left(-\dfrac{1}{2}\right) \\[5ex] \alpha = 120^\circ, 240^\circ ... Unit\:\: Circle\:\: Trig \\[3ex] \rightarrow \cos\alpha = 1 \\[3ex] \alpha = \cos^{-1}(1) \\[3ex] \alpha = 0^\circ, 360^\circ ... Unit\:\: Circle\:\: Trig \\[3ex]$ Check
 $\underline{LHS} \\[3ex] \cos\alpha - \sqrt{3}\sin\alpha \\[3ex] \alpha = 120 \\[5ex] \cos\alpha = \cos 120 = -\dfrac{1}{2} \\[5ex] \sin\alpha = \sin 120 = \dfrac{\sqrt{3}}{2} \\[5ex] \sqrt{3}\sin\alpha = \sqrt{3} * \dfrac{\sqrt{3}}{2} = \dfrac{3}{2} \\[5ex] = -\dfrac{1}{2} - \dfrac{3}{2} \\[5ex] = -\dfrac{1}{2} - \dfrac{3}{2} \\[5ex] = \dfrac{-1 - 3}{2} \\[5ex] = -\dfrac{4}{2} \\[5ex] = -2 \\[3ex] Extraneous\:\ root \\[3ex] NO \\[3ex]$ $\alpha = 240 \\[5ex] \cos\alpha = \cos 240 = -\dfrac{1}{2} \\[5ex] \sin\alpha = \sin 240 = -\dfrac{\sqrt{3}}{2} \\[5ex] \sqrt{3}\sin\alpha = \sqrt{3} * -\dfrac{\sqrt{3}}{2} = -\dfrac{3}{2} \\[5ex] = -\dfrac{1}{2} - - \dfrac{3}{2} \\[5ex] = -\dfrac{1}{2} + \dfrac{3}{2} \\[5ex] = \dfrac{-1 + 3}{2} \\[5ex] = -\dfrac{2}{2} \\[5ex] = 1 \\[3ex] YES \\[3ex]$ $\alpha = 0 \\[3ex] \cos\alpha = \cos 0 = 1 \\[3ex] \sin\alpha = \sin 0 = 0 \\[3ex] \sqrt{3}\sin\alpha = \sqrt{3} * 0 = 0\\[3ex] = 1 - 0 \\[5ex] = 1 \\[3ex] YES \\[3ex]$ $\alpha = 360 \\[3ex] \cos\alpha = \cos 360 = 1 \\[3ex] \sin\alpha = \sin 360 = 0 \\[3ex] \sqrt{3}\sin\alpha = \sqrt{3} * 0 = 0\\[3ex] = 1 - 0 \\[5ex] = 1 \\[3ex] YES$ $\underline{RHS} \\[3ex] 1$

$\underline{Specific\:\: solutions} \\[3ex] \alpha = 240^\circ = 240 * \dfrac{\pi}{180} = \dfrac{4\pi}{3} \\[5ex] \alpha = 0^\circ = 120 * \dfrac{\pi}{180} = 0 \\[5ex] \alpha = 360^\circ = 360 * 120 * \dfrac{\pi}{180} = 2\pi \\[5ex] \underline{General\:\: solutions} \\[3ex] \alpha = 0 + 2\pi k = 2\pi k ... coterminal\:\: \angle s \\[3ex] \alpha = \dfrac{4\pi}{3} + 2\pi k ... coterminal\:\: \angle s$
(11.) Determine all the solutions of $2\cos^2\theta + 3\cos\theta = -1$ in the interval $[0, 2\pi]$

$2\cos^2\theta + 3\cos\theta = -1 \\[3ex] Let\:\: \cos\theta = p \\[3ex] 2p^2 + 3p = -1 \\[3ex] 2p^2 + 3p + 1 = 0 \\[3ex] 2p^2 + 2p + p + 1 = 0 \\[3ex] 2p(p + 1) + 1(p + 1) = 0 \\[3ex] p + 1 = 0 \:\:\:OR\:\:\: 2p + 1 = 0 \\[3ex] p = -1 \:\:\:OR\:\:\: 2p = -1 \\[3ex] p = -1 \:\:\:OR\:\:\: p = -\dfrac{1}{2} \\[5ex] Substitute\:\: back \\[3ex] \rightarrow \cos\theta = -1 \theta = \cos^{-1}(-1) \\[3ex] \theta = 180^\circ ... Unit\:\: Circle\:\: Trig \\[3ex] \rightarrow \cos\theta = -\dfrac{1}{2} \\[5ex] \theta = \cos^{-1}\left(-\dfrac{1}{2}\right) \\[5ex] \theta = 120^\circ, 240^\circ ... Unit\:\: Circle\:\: Trig \\[3ex]$ Check
 $\underline{LHS} \\[3ex] 2\cos^2\theta + 3\cos\theta \\[3ex] \theta = 180 \\[3ex] \cos\theta = \cos 180 = -1 \\[3ex] \cos^2\theta = (\cos 180)^2 = (-1)^2 = 1 \\[3ex] = 2(1) + 3(-1) \\[3ex] = 2 - 3 \\[3ex] = -1 \\[3ex] YES \\[3ex]$ $\theta = 120 \\[3ex] \cos\theta = \cos 120 = -\dfrac{1}{2} \\[3ex] \cos^2\theta = (\cos 120)^2 = \left(-\dfrac{1}{2}\right)^2 = \dfrac{1}{4} \\[3ex] = 2\left(\dfrac{1}{4}\right) + 3\left(-\dfrac{1}{2}\right) \\[3ex] = \dfrac{1}{2} - \dfrac{3}{2} \\[3ex] = \dfrac{1 - 3}{2} \\[5ex] = -\dfrac{2}{2} \\[5ex] = -1 \\[3ex] YES \\[3ex]$ $\theta = 240 \\[3ex] \cos\theta = \cos 240 = -\dfrac{1}{2} \\[3ex] \cos^2\theta = (\cos 240)^2 = \left(-\dfrac{1}{2}\right)^2 = \dfrac{1}{4} \\[3ex] = 2\left(\dfrac{1}{4}\right) + 3\left(-\dfrac{1}{2}\right) \\[3ex] = \dfrac{1}{2} - \dfrac{3}{2} \\[3ex] = \dfrac{1 - 3}{2} \\[5ex] = -\dfrac{2}{2} \\[5ex] = -1 \\[3ex] YES$ $\underline{RHS} \\[3ex] -1$

$\underline{Specific\:\: solutions} \\[3ex] \alpha = 180^\circ = \pi \\[5ex] \alpha = 120^\circ = 120 * \dfrac{\pi}{180} = \dfrac{2\pi}{3} \\[5ex] \alpha = 240^\circ = 240 * \dfrac{\pi}{180} = \dfrac{4\pi}{3} \\[5ex] \underline{General\:\: solutions} \\[3ex] \alpha = \pi + 2\pi k ... coterminal\:\: \angle s \\[3ex] \alpha = \dfrac{2\pi}{3} + 2\pi k ... coterminal\:\: \angle s \\[5ex] \alpha = \dfrac{4\pi}{3} + 2\pi k ... coterminal\:\: \angle s$
(12.) Determine all the solutions of $\sqrt{2}\sin\theta = 2\sin^2\theta$ in the interval $[0, 2\pi)$

$\sqrt{2}\sin\theta = 2\sin^2\theta \\[3ex] 2\sin^2\theta = \sqrt{2}\sin\theta \\[3ex] 2\sin^2\theta - \sqrt{2}\sin\theta = 0 \\[3ex] Let\:\: \sin\theta = p \\[3ex] 2p^2 - p\sqrt{2} = 0 \\[3ex] p(2p - \sqrt{2}) = 0 \\[3ex] p = 0 \:\:\:OR\:\:\: 2p - \sqrt{2} = 0 \\[3ex] p = 0 \:\:\:OR\:\:\: 2p = \sqrt{2} \\[3ex] p = 0 \:\:\:OR\:\:\: p = \dfrac{\sqrt{2}}{2} \\[5ex] Substitute\:\: back \\[3ex] \rightarrow \sin\theta = 0 \theta = \sin^{-1}(0) \\[3ex] \theta = 0^\circ, 180^\circ ... Unit\:\: Circle\:\: Trig \\[3ex] \rightarrow \sin\theta = \dfrac{\sqrt{2}}{2} \\[5ex] \theta = \sin^{-1}\left(\dfrac{\sqrt{2}}{2}\right) \\[5ex] \theta = 45^\circ, 135^\circ ... Unit\:\: Circle\:\: Trig \\[3ex]$ Check
 $\underline{LHS} \\[3ex] \sqrt{2}\sin\theta \\[3ex] \theta = 0 \\[5ex] \sin\theta = \sin 0 = 0 \\[3ex] = \sqrt{2} * 0 \\[5ex] = 0 \\[3ex] YES \\[3ex]$ $\theta = 180 \\[5ex] \sin\theta = \sin 180 = 0 \\[3ex] = \sqrt{2} * 0 \\[5ex] = 0 \\[3ex] YES \\[3ex]$ $\theta = 45 \\[5ex] \sin\theta = \sin 45 = \dfrac{\sqrt{2}}{2} \\[5ex] = \sqrt{2} * \dfrac{\sqrt{2}}{2} \\[5ex] = \dfrac{2}{2} \\[5ex] = 1 \\[3ex] YES \\[3ex]$ $\theta = 135 \\[5ex] \sin\theta = \sin 135 = \dfrac{\sqrt{2}}{2} \\[5ex] = \sqrt{2} * \dfrac{\sqrt{2}}{2} \\[5ex] = \dfrac{2}{2} \\[5ex] = 1 \\[3ex] YES$ $\underline{RHS} \\[3ex] 2\sin^2\theta \\[3ex] \theta = 0 \\[5ex] \sin\theta = \sin 0 = 0 \\[3ex] \sin^2\theta = (\sin\theta)^2 = 0^2 = 0 \\[3ex] = 2 * 0 \\[3ex] = 0 \\[3ex] YES \\[3ex]$ $\theta = 180 \\[5ex] \sin\theta = \sin 180 = 0 \\[3ex] \sin^2\theta = (\sin\theta)^2 = 0^2 = 0 \\[3ex] = 2 * 0 \\[3ex] = 0 \\[3ex] YES \\[3ex]$ $\theta = 45 \\[5ex] \sin\theta = \sin 45 = \dfrac{\sqrt{2}}{2} \\[3ex] \sin^2\theta = (\sin\theta)^2 \\[3ex] \sin^2\theta = \left(\dfrac{\sqrt{2}}{2} \right) \\[5ex] \sin^2\theta = \dfrac{(\sqrt{2})^2}{2^2} \\[5ex] \sin^2\theta = \dfrac{2}{4} = \dfrac{1}{2} \\[5ex] = 2 * \dfrac{1}{2} \\[3ex] = 1 \\[3ex] YES \\[3ex]$ $\theta = 135 \\[5ex] \sin\theta = \sin 135 = \dfrac{\sqrt{2}}{2} \\[3ex] \sin^2\theta = (\sin\theta)^2 \\[3ex] \sin^2\theta = \left(\dfrac{\sqrt{2}}{2} \right) \\[5ex] \sin^2\theta = \dfrac{(\sqrt{2})^2}{2^2} \\[5ex] \sin^2\theta = \dfrac{2}{4} = \dfrac{1}{2} \\[5ex] = 2 * \dfrac{1}{2} \\[3ex] = 1 \\[3ex] YES$

$\underline{Specific\:\: solutions} \\[3ex] \alpha = 0^\circ = 0 * \dfrac{\pi}{180} = 0 \\[5ex] \alpha = 45^\circ = 45 * \dfrac{\pi}{180} = \dfrac{\pi}{4} \\[5ex] \alpha = 135^\circ = 135 * \dfrac{\pi}{180} = \dfrac{3\pi}{4} \\[5ex] \alpha = 180^\circ = \pi \\[5ex] \underline{General\:\: solutions} \\[3ex] \alpha = 0 + 2\pi k = 2\pi k ... coterminal\:\: \angle s \\[3ex] \alpha = \dfrac{\pi}{4} + 2\pi k ... coterminal\:\: \angle s \\[5ex] \alpha = \dfrac{3\pi}{4} + 2\pi k ... coterminal\:\: \angle s \\[5ex] \alpha = \pi + 2\pi k ... coterminal\:\: \angle s$
(13.) Determine all the solutions of $6\cos\beta - 6\sin\beta = 3\sqrt{6}$ in the interval $[0, 2\pi)$

This is not straightforward. It has two functions - the cosine function and the sine function.
What do we do to express it in terms of only one function?

$6\cos\beta - 6\sin\beta = 3\sqrt{6} \\[3ex] Divide\:\: each\:\: term\:\: by\:\: 3 ...Simplify \\[3ex] 2\cos\beta - 2\sin\beta = \sqrt{6} \\[3ex] 2(\cos\beta - \sin\beta) = \sqrt{6} \\[3ex] Square\:\: both\:\: sides \\[3ex] 2^2 * (\cos\beta - \sin\beta)^2 = (\sqrt{6})^2 \\[3ex] 4[(\cos\beta - \sin\beta)(\cos\beta - \sin\beta)] = 6 \\[3ex] 4(\cos^2\beta - \sin\beta\cos\beta - \sin\beta\cos\beta + \sin^2\beta) = 6 \\[3ex] 4(\cos^2\beta + \sin^2\beta - 2\sin\beta\cos\beta) = 6 \\[3ex] \cos^2\beta + \sin^2\beta = 1 ... Pythagorean\:\: Identity \\[3ex] 2\sin\beta\cos\beta = \sin2\beta ... Double\:\: Angle\:\: Formula \\[3ex] 4(1 - 2\sin 2\beta) = 6 \\[3ex] 4 - 4\sin 2\beta = 6 \\[3ex] 4 - 6 = 4\sin 2\beta \\[3ex] -2 = 4\sin 2\beta \\[3ex] 4\sin 2\beta = -2 \\[3ex] \sin2\beta = -\dfrac{2}{4} \\[5ex] \sin2\beta = -\dfrac{1}{2} \\[5ex] Argument = 2\beta \\[3ex] \therefore interval = [0, 4\pi) \\[3ex] 2\beta = \sin^{-1}\left(-\dfrac{1}{2}\right) \\[5ex] \sin^{-1}\left(-\dfrac{1}{2}\right) = 210^\circ, 330^\circ ... Unit\:\: Circle\:\: Trig \\[3ex] Based\:\: on\:\: [0, 4\pi) = [0, 4 * 180^\circ) = [0, 720^\circ) \\[3ex] \sin^{-1}(1) = 210 + 360(1) ...coterminal \angle s \\[3ex] \sin^{-1}(1) = 210 + 360 \\[3ex] \sin^{-1}(1) = 570^\circ \\[3ex] Also,\:\: \sin^{-1}(1) = 330 + 360(1) ...coterminal \angle s \\[3ex] \sin^{-1}(1) = 330 + 360 \\[3ex] \sin^{-1}(1) = 690^\circ \\[3ex] \therefore 2\beta = 210, \:\:2\beta = 330, \:\:2\beta = 570, \:\:2\beta = 690 \\[3ex] \beta = \dfrac{210}{2}, \:\:\beta = \dfrac{330}{2}, \:\:\beta = \dfrac{570}{2}, \:\:\beta = \dfrac{690}{2} \\[5ex] \beta = 105^\circ, 165^\circ, 285^\circ, 345^\circ \\[3ex] 105^\circ = 105 * \dfrac{\pi}{180} = \dfrac{7\pi}{12} \\[5ex] 165^\circ = 165 * \dfrac{\pi}{180} = \dfrac{11\pi}{12} \\[5ex] 285^\circ = 285 * \dfrac{\pi}{180} = \dfrac{19\pi}{12} \\[5ex] 345^\circ = 345 * \dfrac{\pi}{180} = \dfrac{23\pi}{12} \\[5ex] \beta = \dfrac{7\pi}{12}, \dfrac{11\pi}{12}, \dfrac{19\pi}{12}, \dfrac{23\pi}{12} \\[5ex]$ Check
 $\underline{LHS} \\[3ex] 6\cos\beta - 6\sin\beta = 6(\cos\beta - \sin\beta) \\[3ex] \beta = 105^\circ \\[3ex] \cos\beta = \cos105 = \dfrac{\sqrt{2} - \sqrt{6}}{4} ... Special\:\: Angles\:\: Kids \\[5ex] \sin\beta = \sin105 = \dfrac{\sqrt{2} + \sqrt{6}}{4} ... Special\:\: Angles\:\: Kids \\[5ex] \cos\beta - \sin\beta = \dfrac{\sqrt{2} - \sqrt{6}}{4} - \left(\dfrac{\sqrt{2} + \sqrt{6}}{4}\right) \\[5ex] \cos\beta - \sin\beta = \dfrac{\sqrt{2} - \sqrt{6} - (\sqrt{2} + \sqrt{6})}{4} \\[5ex] \cos\beta - \sin\beta = \dfrac{\sqrt{2} - \sqrt{6} - \sqrt{2} - \sqrt{6}}{4} \\[5ex] \cos\beta - \sin\beta = \dfrac{-2\sqrt{6}}{4} = \dfrac{-\sqrt{6}}{2} \\[5ex] = 6\left(-\dfrac{\sqrt{6}}{2}\right) \\[5ex] = 3 * -\sqrt{6} \\[3ex] = -3\sqrt{6} \\[3ex] Extraneous\:\: root \\[3ex] NO \\[3ex]$ $\beta = 165^\circ \\[3ex] \cos\beta = \cos165 = \dfrac{-\sqrt{6} - \sqrt{2}}{4} ... Special\:\: Angles\:\: Kids \\[5ex] \sin\beta = \sin165 = \dfrac{\sqrt{6} - \sqrt{2}}{4} ... Special\:\: Angles\:\: Kids \\[5ex] \cos\beta - \sin\beta = \dfrac{-\sqrt{6} - \sqrt{2}}{4} - \left(\dfrac{\sqrt{6} - \sqrt{2}}{4}\right) \\[5ex] \cos\beta - \sin\beta = \dfrac{-\sqrt{6} - \sqrt{2} - (\sqrt{6} - \sqrt{2})}{4} \\[5ex] \cos\beta - \sin\beta = \dfrac{-\sqrt{6} - \sqrt{2} - \sqrt{6} + \sqrt{2}}{4} \\[5ex] \cos\beta - \sin\beta = \dfrac{-2\sqrt{6}}{4} = \dfrac{-\sqrt{6}}{2} \\[5ex] = 6\left(-\dfrac{\sqrt{6}}{2}\right) \\[5ex] = 3 * -\sqrt{6} \\[3ex] = -3\sqrt{6} \\[3ex] Extraneous\:\: root \\[3ex] NO \\[3ex]$ $\beta = 285^\circ \\[3ex] \cos\beta = \cos285 = \dfrac{\sqrt{6} - \sqrt{2}}{4} ... Special\:\: Angles\:\: Kids \\[5ex] \sin\beta = \sin285 = \dfrac{-\sqrt{2} - \sqrt{6}}{4} ... Special\:\: Angles\:\: Kids \\[5ex] \cos\beta - \sin\beta = \dfrac{\sqrt{6} - \sqrt{2}}{4} - \left(\dfrac{-\sqrt{2} - \sqrt{6}}{4}\right) \\[5ex] \cos\beta - \sin\beta = \dfrac{\sqrt{6} - \sqrt{2} - (-\sqrt{2} - \sqrt{6})}{4} \\[5ex] \cos\beta - \sin\beta = \dfrac{\sqrt{6} - \sqrt{2} + \sqrt{2} + \sqrt{6}}{4} \\[5ex] \cos\beta - \sin\beta = \dfrac{2\sqrt{6}}{4} = \dfrac{\sqrt{6}}{2} \\[5ex] = 6\left(\dfrac{\sqrt{6}}{2}\right) \\[5ex] = 3 * \sqrt{6} \\[3ex] = 3\sqrt{6} \\[3ex] YES \\[3ex]$ $\beta = 345^\circ \\[3ex] \cos\beta = \cos345 = \dfrac{\sqrt{2} + \sqrt{6}}{4} ... Special\:\: Angles\:\: Kids \\[5ex] \sin\beta = \sin345 = \dfrac{\sqrt{2} - \sqrt{6}}{4} ... Special\:\: Angles\:\: Kids \\[5ex] \cos\beta - \sin\beta = \dfrac{\sqrt{2} + \sqrt{6}}{4} - \left(\dfrac{\sqrt{2} - \sqrt{6}}{4}\right) \\[5ex] \cos\beta - \sin\beta = \dfrac{\sqrt{2} + \sqrt{6} - (\sqrt{2} - \sqrt{6})}{4} \\[5ex] \cos\beta - \sin\beta = \dfrac{\sqrt{2} + \sqrt{6} - \sqrt{2} + \sqrt{6}}{4} \\[5ex] \cos\beta - \sin\beta = \dfrac{2\sqrt{6}}{4} = \dfrac{\sqrt{6}}{2} \\[5ex] = 6\left(\dfrac{\sqrt{6}}{2}\right) \\[5ex] = 3 * \sqrt{6} \\[3ex] = 3\sqrt{6} \\[3ex] YES$ $\underline{RHS} \\[3ex] 3\sqrt{6}$

$\underline{Specific\:\: solutions} \\[3ex] \alpha = 285^\circ = \dfrac{19\pi}{12} \\[5ex] \alpha = 345^\circ = \dfrac{23\pi}{12} \\[5ex] \underline{General\:\: solutions} \\[3ex] \alpha = \dfrac{19\pi}{12} + 2\pi k ...coterminal\:\: \angle s \\[5ex] \alpha = \dfrac{23\pi}{12} + 2\pi k ...coterminal\:\: \angle s$
(14.) Determine all the solutions of $\cos\alpha + \sqrt{3}\sin\alpha = 1$ in the interval $[0, 2\pi]$

This is not straightforward. It has two functions - the cosine function and the sine function.
What do we do to express it in terms of only one function?

$\cos\alpha + \sqrt{3}\sin\alpha = 1 \\[3ex] \sqrt{3}\sin\alpha = 1 - \cos\alpha \\[3ex] Square\:\: both\:\: sides \\[3ex] (\sqrt{3}\sin\alpha)^2 = (1 - \cos\alpha)^2 \\[3ex] (\sqrt{3})^2 * \sin^2\alpha = (1 - \cos\alpha)(1 - \cos\alpha) \\[3ex] 3\sin^2\alpha = 1 - \cos\alpha - \cos\alpha + \cos^2\alpha \\[3ex] 3\sin^2\alpha = 1 - 2\cos\alpha + \cos^2\alpha \\[3ex] \sin^2\alpha + \cos^2\alpha = 1 ... Pythagorean\:\: Identity \\[3ex] \sin^2\alpha = 1 - \cos^2\alpha \\[3ex] 3(1 - \cos^2\alpha) = 1 - 2\cos\alpha + \cos^2\alpha \\[3ex] 3 - 3\cos^2\alpha = 1 - 2\cos\alpha + \cos^2\alpha \\[3ex] 0 = 1 - 2\cos\alpha + \cos^2\alpha - 3 + 3\cos^2\alpha \\[3ex] 1 - 2\cos\alpha + \cos^2\alpha - 3 + 3\cos^2\alpha = 0 \\[3ex] 4\cos^2\alpha - 2\cos\alpha - 2 = 0 \\[3ex] Divide\:\: all\:\: terms\:\: by\:\: 2 \\[3ex] 2\cos^2\alpha - \cos\alpha - 1 = 0 \\[3ex] Let\:\: \cos\alpha = p \\[3ex] 2p^2 - p - 1 = 0 \\[3ex] 2p^2 + p - 2p - 1 = 0 \\[3ex] p(2p + 1) - 1(2p + 1) = 0 \\[3ex] (2p + 1)(p - 1) = 0 \\[3ex] 2p + 1 = 0 \:\:\:OR\:\:\: p - 1 = 0 \\[3ex] 2p = -1 \:\:\:OR\:\:\: p = 1 \\[3ex] p = -\dfrac{1}{2} \:\:\:OR\:\:\: p = 1 \\[3ex] Substitute\:\: back \\[3ex] \rightarrow \cos\alpha = -\dfrac{1}{2} \\[3ex] \alpha = \cos^{-1}\left(-\dfrac{1}{2}\right) \\[5ex] \alpha = 120^\circ, 240^\circ ... Unit\:\: Circle\:\: Trig \\[3ex] \rightarrow \cos\alpha = 1 \\[3ex] \alpha = \cos^{-1}(1) \\[3ex] \alpha = 0^\circ, 360^\circ ... Unit\:\: Circle\:\: Trig \\[3ex]$ Check
 $\underline{LHS} \\[3ex] \cos\alpha + \sqrt{3}\sin\alpha \\[3ex] \alpha = 120 \\[5ex] \cos\alpha = \cos 120 = -\dfrac{1}{2} \\[5ex] \sin\alpha = \sin 120 = \dfrac{\sqrt{3}}{2} \\[5ex] \sqrt{3}\sin\alpha = \sqrt{3} * \dfrac{\sqrt{3}}{2} = \dfrac{3}{2} \\[5ex] = -\dfrac{1}{2} + \dfrac{3}{2} \\[5ex] = -\dfrac{1}{2} + \dfrac{3}{2} \\[5ex] = \dfrac{-1 + 3}{2} \\[5ex] = -\dfrac{2}{2} \\[5ex] = 1 \\[3ex] YES \\[3ex]$ $\alpha = 240 \\[5ex] \cos\alpha = \cos 240 = -\dfrac{1}{2} \\[5ex] \sin\alpha = \sin 240 = -\dfrac{\sqrt{3}}{2} \\[5ex] \sqrt{3}\sin\alpha = \sqrt{3} * -\dfrac{\sqrt{3}}{2} = -\dfrac{3}{2} \\[5ex] = -\dfrac{1}{2} + - \dfrac{3}{2} \\[5ex] = -\dfrac{1}{2} - \dfrac{3}{2} \\[5ex] = \dfrac{-1 - 3}{2} \\[5ex] = -\dfrac{4}{2} \\[5ex] = -2 \\[3ex] Extraneous\:\: root \\[3ex] NO \\[3ex]$ $\alpha = 0 \\[3ex] \cos\alpha = \cos 0 = 1 \\[3ex] \sin\alpha = \sin 0 = 0 \\[3ex] \sqrt{3}\sin\alpha = \sqrt{3} * 0 = 0\\[3ex] = 1 + 0 \\[5ex] = 1 \\[3ex] YES \\[3ex]$ $\alpha = 360 \\[3ex] \cos\alpha = \cos 360 = 1 \\[3ex] \sin\alpha = \sin 360 = 0 \\[3ex] \sqrt{3}\sin\alpha = \sqrt{3} * 0 = 0\\[3ex] = 1 + 0 \\[5ex] = 1 \\[3ex] YES$ $\underline{RHS} \\[3ex] 1$

$\underline{Specific\:\: solutions} \\[3ex] \alpha = 120^\circ = 120 * \dfrac{\pi}{180} = \dfrac{2\pi}{3} \\[5ex] \alpha = 0^\circ = 120 * \dfrac{\pi}{180} = 0 \\[5ex] \alpha = 360^\circ = 360 * 120 * \dfrac{\pi}{180} = 2\pi \\[5ex] \underline{General\:\: solutions} \\[3ex] \alpha = 0 + 2\pi k = 2\pi k ... coterminal\:\: \angle s \\[3ex] \alpha = \dfrac{2\pi}{3} + 2\pi k ... coterminal\:\: \angle s$
(15.) Determine all the solutions of $\cos(\pi + \theta) + \sin\left(\theta - \dfrac{\pi}{2}\right) = 1$ in the interval $[0, 2\pi)$

This is not straightforward. It has two functions - the cosine function and the sine function.
How do we express it in terms of only one function?
Let us apply the Sum and Difference Formulas first

$\cos(\pi + \theta) + \sin\left(\theta - \dfrac{\pi}{2}\right) = 1 \\[5ex] \cos(\pi + \theta) = \cos\pi\cos\theta - \sin\pi\sin\theta ... Addition\:\: Formula \\[3ex] \cos\pi = -1 \:\:and\:\: \sin\pi = 0 ... Unit\:\: Circle\:\: Trig \\[3ex] \rightarrow \cos(\pi + \theta) = -1 * \cos\theta - 0 * \sin\theta = -\cos\theta - 0 = -\cos\theta \\[3ex] \sin\left(\theta - \dfrac{\pi}{2}\right) = \sin\theta\cos\dfrac{\pi}{2} - \cos\theta\sin\dfrac{\pi}{2} ... Difference\:\: Formula \\[5ex] \cos\dfrac{\pi}{2} = 0 \:\:and\:\: \sin\dfrac{\pi}{2} = 1 ... Unit\:\: Circle\:\: Trig \\[5ex] \rightarrow \sin\left(\theta - \dfrac{\pi}{2}\right) = \sin\theta * 0 - \cos\theta * 1 = 0 - \cos\theta = -\cos\theta \\[5ex] \cos(\pi + \theta) + \sin\left(\theta - \dfrac{\pi}{2}\right) = 1 \\[5ex] \rightarrow -\cos\theta + -\cos\theta = 1 \\[3ex] -\cos\theta - \cos\theta = 1 \\[3ex] -2\cos\theta = 1 \\[5ex] \cos\theta = -\dfrac{1}{2} \\[5ex] \theta = \cos^{-1}{-\dfrac{1}{2}} \\[5ex] \theta = 120^\circ, 240^\circ \\[3ex]$ Check
 $\underline{LHS} \\[3ex] \cos(\pi + \theta) + \sin\left(\theta - \dfrac{\pi}{2}\right) \\[5ex] \pi = 180^\circ \\[3ex] \dfrac{\pi}{2} = \dfrac{180}{2} = 90^circ \\[5ex] \theta = 120^\circ \\[3ex] = \cos(180 + 120) + \sin(120 - 90) \\[3ex] = \cos 300 + \sin 30 \\[3ex] = \dfrac{1}{2} + \dfrac{1}{2} ... Unit\:\: Circle\:\: Trig \\[5ex] = 1 \\[3ex] YES \\[3ex]$ $\theta = 240^\circ \\[3ex] = \cos(180 + 240) + \sin(240 - 90) \\[3ex] = \cos 420 + \sin 150 \\[3ex] coterminal\:\: \angle \:\:of\:\: 420 = 420 - 360 = 60 \\[3ex] = \cos 60 + \sin 150 \\[3ex] = \dfrac{1}{2} + \dfrac{1}{2} ... Unit\:\: Circle\:\: Trig \\[5ex] = 1 \\[3ex] YES$ $\underline{RHS} \\[3ex] 1$

$\underline{Specific\:\: solutions} \\[3ex] \beta = 120^\circ = 120 * \dfrac{\pi}{180} = \dfrac{2\pi}{3} \\[5ex] \beta = 240^\circ = 240 * \dfrac{\pi}{180} = \dfrac{4\pi}{3} \\[5ex] \underline{General\:\: solutions} \\[3ex] \beta = \dfrac{2\pi}{3} + 2\pi k ... coterminal\:\: \angle s \\[5ex] \beta = \dfrac{4\pi}{3} + 2\pi k ... coterminal\:\: \angle s$
(16.) Determine all the solutions of $\cot\beta + \sqrt{3} = \csc\beta$ in the interval $0 \le \beta \lt 2\pi$

Interval of $0 \le \beta \lt 2\pi$ means $[0, 2\pi)$
This is not straightforward. It has two functions - the cotangent function and the cosecant function.
How do we express it in terms of only one function?

$\cot\beta + \sqrt{3} = \csc\beta \\[3ex] Square\:\: both\:\: sides \\[3ex] (\cot\beta + \sqrt{3})^2 = (\csc\beta)^2 \\[3ex] (\cot\beta + \sqrt{3})(\cot\beta + \sqrt{3}) = \csc^2\beta \\[3ex] \cot^2\beta + \sqrt{3}\cot\beta + \sqrt{3}\cot\beta + (\sqrt{3})^2 = \csc^2\beta \\[3ex] \cot^2\beta + 2\sqrt{3}\cot\beta + 3 = \csc^2\beta \\[3ex] \csc^2\beta = 1 + \cot^2\beta ... Pythagorean\:\: Identity \\[3ex] \cot^2\beta + 2\sqrt{3}\cot\beta + 3 = 1 + \cot^2\beta \\[3ex] \cot^2\beta - \cot^2\beta + 2\srt{3}\cot\beta + 3 - 1 = 0 \\[3ex] 2\sqrt{3}\cot\beta + 2 = 0 \\[3ex] 2\sqrt{3}\cot\beta = -2 \\[3ex] \cot\beta = -\dfrac{2}{2\sqrt{3}} \\[5ex] \cot\beta = -\dfrac{1}{\sqrt{3}} \\[5ex] \cot\beta = -\dfrac{1}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\[5ex] \cot\beta = -\dfrac{\sqrt{3}}{3} \\[5ex] \beta = \cos^{-1}\left({-\dfrac{\sqrt{3}}{3}}\right) \\[5ex] \beta = 120^\circ, 300^\circ ... Unit\:\: Circle\:\: Trig \\[3ex]$ Check
 $\underline{LHS} \\[3ex] \cot\beta + \sqrt{3} \\[3ex] \beta = 120^\circ \\[5ex] \cot\beta = \cot 120 = -\dfrac{\sqrt{3}}{3} \\[5ex] = -\dfrac{\sqrt{3}}{3} + \sqrt{3} \\[5ex] = -\dfrac{\sqrt{3}}{3} + \dfrac{3\sqrt{3}}{3} \\[5ex] = \dfrac{-\sqrt{3} + 3\sqrt{3}}{3} \\[5ex] = \dfrac{2\sqrt{3}}{3} \\[3ex] YES \\[3ex]$ $\beta = 300^\circ \\[5ex] \cot\beta = \cot 300 = -\dfrac{\sqrt{3}}{3} \\[5ex] = -\dfrac{\sqrt{3}}{3} + \sqrt{3} \\[5ex] = -\dfrac{\sqrt{3}}{3} + \dfrac{3\sqrt{3}}{3} \\[5ex] = \dfrac{-\sqrt{3} + 3\sqrt{3}}{3} \\[5ex] = \dfrac{2\sqrt{3}}{3} \\[3ex] LHS \ne RHS \\[3ex] NO$ $\underline{RHS} \\[3ex] \csc\beta \\[3ex] \beta = 120^\circ \\[3ex] = \csc 120 \\[3ex] = \dfrac{2\sqrt{3}}{3} \\[3ex] YES \\[3ex]$ $\beta = 300^\circ \\[3ex] = \csc 300 \\[3ex] = -\dfrac{2\sqrt{3}}{3} \\[5ex] RHS \ne LHS \\[3ex] NO$

$\underline{Specific\:\: solution} \\[3ex] \beta = 120^\circ = 120 * \dfrac{\pi}{180} = \dfrac{2\pi}{3} \\[5ex] \underline{General\:\: solution} \\[3ex] \beta = \dfrac{2\pi}{3} + 2\pi k ... coterminal\:\: \angle s$
(17.) Determine all the solutions of $\cos(2\alpha) + 14\sin^2\alpha = 10$ in the interval $[0, 2\pi)$

This is not straightforward. It has two functions - the cosine function and the sine function.
How do we express it in terms of only one function?
Let us apply the Sum and Difference Formulas first

$\cos(2\alpha) + 14\sin^2\alpha = 10 \\[3ex] \cos(2\alpha) = 1 - 2\sin^2\alpha ... Double-Angle\:\: Formula \\[3ex] \rightarrow 1 - 2\sin^2\alpha + 14\sin^2\alpha = 10 \\[3ex] 1 + 12\sin^2\alpha = 10 \\[3ex] 12\sin^2\alpha = 10 - 1 \\[3ex] 12\sin^2\alpha = 9 \\[3ex] \sin^2\alpha = \dfrac{9}{12} = \dfrac{3}{4} \\[5ex] \sin\alpha = \pm \sqrt{\dfrac{3}{4}} = \pm \dfrac{\sqrt{3}}{2} \\[5ex] \alpha = \sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right) \:\:OR\:\: \alpha = \sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right) \\[5ex] \alpha = 60, 120 \:\:OR\:\: \alpha = 240, 300 \\[3ex] \alpha = 60^\circ, 120^\circ, 240^\circ, 300^\circ ... Unit\:\: Circle\:\: Trig \\[3ex]$ Check
 $\underline{LHS} \\[3ex] \cos(2\alpha) + 14\sin^2\alpha \\[3ex] \alpha = 60^\circ \\[3ex] 2\alpha = 2 * 60 = 120 \\[3ex] \cos2\alpha = \cos 120 = -\dfrac{1}{2} \\[5ex] \sin^2\alpha = (\sin 60)^2 = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{(\sqrt{3})^2}{2^2} = \dfrac{3}{4} \\[5ex] 14\sin^2\alpha = 14\left(\dfrac{3}{4}\right) = 7\left(\dfrac{3}{2}\right) = \dfrac{21}{2} \\[5ex] = -\dfrac{1}{2} + \dfrac{21}{2} \\[5ex] = \dfrac{-1 + 21}{2} \\[5ex] = \dfrac{20}{2} \\[5ex] = 10 \\[3ex] YES \\[3ex]$ $\alpha = 120^\circ \\[3ex] 2\alpha = 2 * 120 = 240 \\[3ex] \cos2\alpha = \cos 240 = -\dfrac{1}{2} \\[5ex] \sin^2\alpha = (\sin 120)^2 = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{(\sqrt{3})^2}{2^2} = \dfrac{3}{4} \\[5ex] 14\sin^2\alpha = 14\left(\dfrac{3}{4}\right) = 7\left(\dfrac{3}{2}\right) = \dfrac{21}{2} \\[5ex] = -\dfrac{1}{2} + \dfrac{21}{2} \\[5ex] = \dfrac{-1 + 21}{2} \\[5ex] = \dfrac{20}{2} \\[5ex] = 10 \\[3ex] YES \\[3ex]$ $\alpha = 240^\circ \\[3ex] 2\alpha = 2 * 240 = 480 \\[3ex] \cos2\alpha = \cos 480 \\[3ex] coterminal\:\: \angle \:\:of\:\: 480 = 480 - 360 = 120 \\[3ex] \cos2\alpha = \cos 480 = \cos 120 = -\dfrac{1}{2} \\[5ex] \sin^2\alpha = (\sin 480)^2 = (\sin 120)^2 = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{(\sqrt{3})^2}{2^2} = \dfrac{3}{4} \\[5ex] 14\sin^2\alpha = 14\left(\dfrac{3}{4}\right) = 7\left(\dfrac{3}{2}\right) = \dfrac{21}{2} \\[5ex] = -\dfrac{1}{2} + \dfrac{21}{2} \\[5ex] = \dfrac{-1 + 21}{2} \\[5ex] = \dfrac{20}{2} \\[5ex] = 10 \\[3ex] YES \\[3ex]$ $\alpha = 300^\circ \\[3ex] 2\alpha = 2 * 300 = 600 \\[3ex] \cos2\alpha = \cos 600 \\[3ex] coterminal\:\: \angle \:\:of\:\: 600 = 600 - 360 = 240 \\[3ex] \cos2\alpha = \cos 600 = \cos 240 = -\dfrac{1}{2} \\[5ex] \sin^2\alpha = (\sin 600)^2 = (\sin 240)^2 = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{(\sqrt{3})^2}{2^2} = \dfrac{3}{4} \\[5ex] 14\sin^2\alpha = 14\left(\dfrac{3}{4}\right) = 7\left(\dfrac{3}{2}\right) = \dfrac{21}{2} \\[5ex] = -\dfrac{1}{2} + \dfrac{21}{2} \\[5ex] = \dfrac{-1 + 21}{2} \\[5ex] = \dfrac{20}{2} \\[5ex] = 10 \\[3ex] YES$ $\underline{RHS} \\[3ex] 10$

$\underline{Specific\:\: solutions} \\[3ex] \alpha = 60^\circ = 60 * \dfrac{\pi}{180} = \dfrac{\pi}{3} \\[5ex] \alpha = 120^\circ = 120 * \dfrac{\pi}{180} = \dfrac{2\pi}{3} \\[5ex] \alpha = 240^\circ = 240 * \dfrac{\pi}{180} = \dfrac{4\pi}{3} \\[5ex] \alpha = 300^\circ = 300 * \dfrac{\pi}{180} = \dfrac{5\pi}{3} \\[5ex] \underline{General\:\: solutions} \\[3ex] \alpha = \dfrac{\pi}{3} + 2\pi k ... coterminal\:\: \angle s \\[5ex] \alpha = \dfrac{2\pi}{3} + 2\pi k ... coterminal\:\: \angle s \\[5ex] \alpha = \dfrac{4\pi}{3} + 2\pi k ... coterminal\:\: \angle s \\[5ex] \alpha = \dfrac{5\pi}{3} + 2\pi k ... coterminal\:\: \angle s$
(18.) Determine all the solutions of $3\tan\alpha\sin\alpha - 3\tan\alpha = 0$ in the interval $[0, 2\pi)$

This is not straightforward. It has two functions: the tangent function and the cosecant function.
How do we express it in terms of only one function?
If we cannot express it in terms of only one function, can we factor by GCF?

$3\tan\alpha\sin\alpha - 3\tan\alpha = 0 \\[3ex] GCF = 3\tan\alpha \\[3ex] Factor\:\: by\:\: GCF \\[3ex] 3\tan\alpha(\sin\alpha - 1) = 0 \\[3ex] 3\tan\alpha = 0 \:\:OR\:\: \sin\alpha - 1 = 0 \\[3ex] \tan\alpha = \dfrac{0}{3} \:\:OR\:\: \sin\alpha = 1 \\[3ex] \tan\alpha = 0 \:\:OR\:\: \sin\alpha = 1 \\[3ex] \alpha = \tan^{-1}(0) \:\:OR\:\: \alpha = \sin^{-1}(1) \\[3ex] \alpha = 0, 180 \:\:OR\:\: \alpha = 90 \\[3ex] \alpha = 0^\circ, 90^\circ, 180^\circ ... Unit\:\: Circle\:\: Trig \\[3ex]$ Check
 $\underline{LHS} \\[3ex] 3\tan\alpha\sin\alpha - 3\tan\alpha \\[3ex] \alpha = 0^\circ \\[5ex] \tan\alpha = \tan 0 = 0 \\[3ex] \sin\alpha = \sin 0 = 0 \\[3ex] = 3 * 0 * 0 - 3 * 0 \\[3ex] = 0 - 0 \\[3ex] = 0 \\[3ex] YES \\[3ex]$ $\alpha = 90^\circ \\[5ex] \tan\alpha = \tan 90 \:\:is\:\: undefined \\[3ex] STOP \\[3ex] NO \\[3ex]$ $\alpha = 180^\circ \\[5ex] \tan\alpha = \tan 180 = 0 \\[3ex] \sin\alpha = \sin 180 = 0 \\[3ex] = 3 * 0 * 0 - 3 * 0 \\[3ex] = 0 - 0 \\[3ex] = 0 \\[3ex] YES$ $\underline{RHS} \\[3ex] 0$

$\underline{Specific\:\: solutions} \\[3ex] \alpha = 0^\circ = 0 * \dfrac{\pi}{180} = 0 \\[5ex] \alpha = 180^\circ = 180 * \dfrac{\pi}{180} = \pi \\[5ex] \underline{General\:\: solutions} \\[3ex] \alpha = 0 + 2\pi k = 2\pi k ... coterminal\:\: \angle s \\[3ex] \alpha = \pi + 2\pi k ... coterminal\:\: \angle s$