1. (a) Prove the identity $\cos 3\theta \equiv 4 \cos^3 \theta - 3 \cos \theta$ by expressing $3\theta$ as $2\theta + \theta$. Use the cosine addition formula: $\cos(a+b) = \cos a \cos b - \sin a \sin b$. So, $\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos \theta - \sin 2\theta \sin \theta$. Recall double angle formulas: $\cos 2\theta = 2 \cos^2 \theta - 1$ and $\sin 2\theta = 2 \sin \theta \cos \theta$. Substitute: $\cos 3\theta = (2 \cos^2 \theta - 1) \cos \theta - 2 \sin \theta \cos \theta \sin \theta$. Simplify: $\cos 3\theta = 2 \cos^3 \theta - \cos \theta - 2 \sin^2 \theta \cos \theta$. Use $\sin^2 \theta = 1 - \cos^2 \theta$: $\cos 3\theta = 2 \cos^3 \theta - \cos \theta - 2 (1 - \cos^2 \theta) \cos \theta$. Expand: $\cos 3\theta = 2 \cos^3 \theta - \cos \theta - 2 \cos \theta + 2 \cos^3 \theta$. Combine like terms: $\cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta$. (b) Solve $\cos 3\theta + \cos \theta \cos 2\theta = \cos^2 \theta$ for $0^\circ \leq \theta \leq 180^\circ$. Rewrite $\cos 3\theta$ using the identity: $4 \cos^3 \theta - 3 \cos \theta + \cos \theta \cos 2\theta = \cos^2 \theta$. Use $\cos 2\theta = 2 \cos^2 \theta - 1$: $4 \cos^3 \theta - 3 \cos \theta + \cos \theta (2 \cos^2 \theta - 1) = \cos^2 \theta$. Expand: $4 \cos^3 \theta - 3 \cos \theta + 2 \cos^3 \theta - \cos \theta = \cos^2 \theta$. Combine terms: $6 \cos^3 \theta - 4 \cos \theta = \cos^2 \theta$. Rearranged: $6 \cos^3 \theta - \cos^2 \theta - 4 \cos \theta = 0$. Let $x = \cos \theta$, then $6 x^3 - x^2 - 4 x = 0$. Factor: $x (6 x^2 - x - 4) = 0$. So, $x=0$ or solve $6 x^2 - x - 4 = 0$. Use quadratic formula: $x = \frac{1 \pm \sqrt{1 + 96}}{12} = \frac{1 \pm \sqrt{97}}{12}$. Since $\cos \theta$ must be between -1 and 1, check values: $\frac{1 + \sqrt{97}}{12} \approx 0.93$ (valid), $\frac{1 - \sqrt{97}}{12} \approx -0.72$ (valid). For $x=0$, $\cos \theta = 0 \Rightarrow \theta = 90^\circ$. For $x \approx 0.93$, $\theta \approx \cos^{-1}(0.93) = 21.8^\circ$ or $338.2^\circ$ (outside range). For $x \approx -0.72$, $\theta \approx \cos^{-1}(-0.72) = 135.7^\circ$. Valid solutions in $0^\circ \leq \theta \leq 180^\circ$ are $21.8^\circ$, $90^\circ$, and $135.7^\circ$. 2. (a) Show $\cot^2 \theta + 2 \cos 2\theta = 4$ can be written as $4 \sin^4 \theta + 3 \sin^2 \theta - 1 = 0$. Recall $\cot \theta = \frac{\cos \theta}{\sin \theta}$, so $\cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta}$. Rewrite equation: $\frac{\cos^2 \theta}{\sin^2 \theta} + 2 \cos 2\theta = 4$. Multiply both sides by $\sin^2 \theta$: $\cos^2 \theta + 2 \sin^2 \theta \cos 2\theta = 4 \sin^2 \theta$. Use $\cos 2\theta = 1 - 2 \sin^2 \theta$: $\cos^2 \theta + 2 \sin^2 \theta (1 - 2 \sin^2 \theta) = 4 \sin^2 \theta$. Expand: $\cos^2 \theta + 2 \sin^2 \theta - 4 \sin^4 \theta = 4 \sin^2 \theta$. Use $\cos^2 \theta = 1 - \sin^2 \theta$: $1 - \sin^2 \theta + 2 \sin^2 \theta - 4 \sin^4 \theta = 4 \sin^2 \theta$. Simplify left: $1 + \sin^2 \theta - 4 \sin^4 \theta = 4 \sin^2 \theta$. Bring all terms to one side: $1 + \sin^2 \theta - 4 \sin^4 \theta - 4 \sin^2 \theta = 0$. Simplify: $1 - 3 \sin^2 \theta - 4 \sin^4 \theta = 0$. Multiply by -1: $4 \sin^4 \theta + 3 \sin^2 \theta - 1 = 0$. (b) Solve $4 \sin^4 \theta + 3 \sin^2 \theta - 1 = 0$ for $0^\circ < \theta < 360^\circ$. Let $x = \sin^2 \theta$, then $4 x^2 + 3 x - 1 = 0$. Use quadratic formula: $x = \frac{-3 \pm \sqrt{9 + 16}}{8} = \frac{-3 \pm 5}{8}$. Solutions: $x = \frac{2}{8} = 0.25$ or $x = \frac{-8}{8} = -1$ (discard negative). So, $\sin^2 \theta = 0.25 \Rightarrow \sin \theta = \pm 0.5$. Find $\theta$ where $\sin \theta = 0.5$: $\theta = 30^\circ, 150^\circ$. Where $\sin \theta = -0.5$: $\theta = 210^\circ, 330^\circ$. 3. (a) Given $\sin(x + \frac{\pi}{6}) - \sin(x - \frac{\pi}{6}) = \cos(x + \frac{\pi}{3}) - \cos(x - \frac{\pi}{3})$, find exact value of $\tan x$. Use sine difference formula: $\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$. Left side: $2 \cos x \sin \frac{\pi}{6} = 2 \cos x \times \frac{1}{2} = \cos x$. Use cosine difference formula: $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$. Right side: $-2 \sin x \sin \frac{\pi}{3} = -2 \sin x \times \frac{\sqrt{3}}{2} = - \sqrt{3} \sin x$. Equation: $\cos x = - \sqrt{3} \sin x$. Divide both sides by $\cos x$: $1 = - \sqrt{3} \tan x$. So, $\tan x = - \frac{1}{\sqrt{3}} = - \frac{\sqrt{3}}{3}$. (b) Find exact roots of the equation for $0 \leq x \leq 2\pi$. $\tan x = - \frac{\sqrt{3}}{3}$. General solution: $x = \arctan(- \frac{\sqrt{3}}{3}) + k \pi$. $\arctan(- \frac{\sqrt{3}}{3}) = - \frac{\pi}{6}$. Within $0 \leq x \leq 2\pi$, solutions are: $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ and $x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$. 4. (a) Show $\sin 2\theta + \cos 2\theta = 2 \sin^2 \theta$ can be expressed as $\cos^2 \theta + 2 \sin \theta \cos \theta - 3 \sin^2 \theta = 0$. Use double angle formulas: $\sin 2\theta = 2 \sin \theta \cos \theta$, $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$. Sum: $2 \sin \theta \cos \theta + \cos^2 \theta - \sin^2 \theta = 2 \sin^2 \theta$. Bring all terms to one side: $\cos^2 \theta + 2 \sin \theta \cos \theta - \sin^2 \theta - 2 \sin^2 \theta = 0$. Simplify: $\cos^2 \theta + 2 \sin \theta \cos \theta - 3 \sin^2 \theta = 0$. (b) Solve $\sin 2\theta + \cos 2\theta = 2 \sin^2 \theta$ for $0^\circ < \theta < 180^\circ$. Rewrite as $\cos^2 \theta + 2 \sin \theta \cos \theta - 3 \sin^2 \theta = 0$. Divide by $\cos^2 \theta$ (assuming $\cos \theta \neq 0$): $1 + 2 \tan \theta - 3 \tan^2 \theta = 0$. Let $t = \tan \theta$, then $-3 t^2 + 2 t + 1 = 0$. Multiply by -1: $3 t^2 - 2 t - 1 = 0$. Use quadratic formula: $t = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm 4}{6}$. Solutions: $t = 1$ or $t = -\frac{1}{3}$. Find $\theta$ where $\tan \theta = 1$: $\theta = 45^\circ, 225^\circ$ (only $45^\circ$ in range). Where $\tan \theta = -\frac{1}{3}$: $\theta = \arctan(-\frac{1}{3}) \approx -18.43^\circ$ or $161.57^\circ$ (in range). Valid solutions: $45^\circ$ and $161.57^\circ$. 5. Solve $2 \cos x - \cos \frac{x}{2} = 1$ for $0 \leq x \leq 2\pi$. Let $y = \frac{x}{2}$, then $x = 2y$. Rewrite: $2 \cos 2y - \cos y = 1$. Use double angle formula: $\cos 2y = 2 \cos^2 y - 1$. Substitute: $2 (2 \cos^2 y - 1) - \cos y = 1$. Expand: $4 \cos^2 y - 2 - \cos y = 1$. Bring all terms to one side: $4 \cos^2 y - \cos y - 3 = 0$. Let $z = \cos y$, then $4 z^2 - z - 3 = 0$. Use quadratic formula: $z = \frac{1 \pm \sqrt{1 + 48}}{8} = \frac{1 \pm 7}{8}$. Solutions: $z = 1$ or $z = -\frac{3}{4}$. For $z=1$, $\cos y = 1 \Rightarrow y = 0$. For $z = -\frac{3}{4}$, $y = \arccos(-0.75) \approx 138.59^\circ$ or $221.41^\circ$. Recall $x = 2y$, so: $x = 0^\circ$, $x = 277.18^\circ$, $x = 442.82^\circ$. Only $0^\circ$ and $277.18^\circ$ are in $0^\circ \leq x \leq 360^\circ$. Convert to radians: $0$, $\approx 4.84$. Final answers: 1. (a) $\cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta$. (b) $\theta = 21.8^\circ, 90^\circ, 135.7^\circ$. 2. (a) $4 \sin^4 \theta + 3 \sin^2 \theta - 1 = 0$. (b) $\theta = 30^\circ, 150^\circ, 210^\circ, 330^\circ$. 3. (a) $\tan x = - \frac{\sqrt{3}}{3}$. (b) $x = \frac{5\pi}{6}, \frac{11\pi}{6}$. 4. (a) $\cos^2 \theta + 2 \sin \theta \cos \theta - 3 \sin^2 \theta = 0$. (b) $\theta = 45^\circ, 161.57^\circ$. 5. $x = 0, 4.84$ radians (approx).