Trigonometric Equations

(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 $[0, 2\pi]$

$ \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 $

(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 $

Solved Examples on Trigonometric Equations