1. Solve the equation $3 \cot x - 4 \cot 2x = 3$ for $0^\circ \leq x \leq 180^\circ$. 2. (a) Express $7 \sin \theta + 24 \cos \theta$ in the form $R \cos (\theta - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{1}{2} \pi$. Give $\alpha$ correct to 4 decimal places. (b) Hence solve $7 \sin \left(\frac{x}{3}\right) + 24 \cos \left(\frac{x}{3}\right) = 24.5$ for $0 < x < \pi$. 3. By expressing $\tan (x - 60^\circ) = 2 \cot x$ as a quadratic in $\tan x$, solve for $0^\circ \leq x \leq 180^\circ$. 4. (a) Show $\tan^3 x + 2 \tan 2x - \tan x = 0$ can be written as $\tan^4 x - 2 \tan^2 x - 3 = 0$ for $\tan x \neq 0$. (b) Hence solve $\tan^3 2\theta + 2 \tan 4\theta - \tan 2\theta = 0$ for $0 < \theta < \pi$. 5. (a) Express $3 \sin x + 2\sqrt{2} \cos \left(x + \frac{\pi}{4}\right)$ as $R \sin (x + \alpha)$, $R > 0$, $0 < \alpha < \frac{\pi}{2}$. State exact $R$ and $\alpha$ to 3 decimals. (b) Solve $6 \sin \left(\frac{\theta}{2}\right) + 4\sqrt{2} \cos \left(\frac{\theta}{2} + \frac{\pi}{4}\right) = 3$ for $-4\pi < \theta < 4\pi$. --- ### Problem 1 1. Given $3 \cot x - 4 \cot 2x = 3$, recall $\cot 2x = \frac{\cot^2 x - 1}{2 \cot x}$. 2. Substitute: $3 \cot x - 4 \cdot \frac{\cot^2 x - 1}{2 \cot x} = 3$. 3. Multiply both sides by $2 \cot x$: $6 \cot^2 x - 4 (\cot^2 x - 1) = 6 \cot x$. 4. Expand: $6 \cot^2 x - 4 \cot^2 x + 4 = 6 \cot x$. 5. Simplify: $2 \cot^2 x + 4 = 6 \cot x$. 6. Rearrange: $2 \cot^2 x - 6 \cot x + 4 = 0$. 7. Divide by 2: $\cot^2 x - 3 \cot x + 2 = 0$. 8. Factor: $(\cot x - 1)(\cot x - 2) = 0$. 9. So $\cot x = 1$ or $\cot x = 2$. 10. For $\cot x = 1$, $x = 45^\circ, 225^\circ$; only $45^\circ$ in $[0^\circ,180^\circ]$. 11. For $\cot x = 2$, $x = \arccot 2 = \arctan \frac{1}{2} \approx 26.565^\circ$. 12. Solutions: $x = 26.565^\circ, 45^\circ$. ### Problem 2 (a) 1. Express $7 \sin \theta + 24 \cos \theta$ as $R \cos (\theta - \alpha)$. 2. Use identity: $R \cos (\theta - \alpha) = R (\cos \theta \cos \alpha + \sin \theta \sin \alpha)$. 3. Equate coefficients: $7 = R \sin \alpha$, $24 = R \cos \alpha$. 4. Find $R = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$. 5. Find $\alpha = \arctan \frac{7}{24} \approx 0.2838$ radians (4 decimals). (b) 1. Given $7 \sin \frac{x}{3} + 24 \cos \frac{x}{3} = 24.5$. 2. Rewrite as $25 \cos \left( \frac{x}{3} - 0.2838 \right) = 24.5$. 3. Divide: $\cos \left( \frac{x}{3} - 0.2838 \right) = \frac{24.5}{25} = 0.98$. 4. Solve: $\frac{x}{3} - 0.2838 = \pm \arccos 0.98 + 2k\pi$. 5. $\arccos 0.98 \approx 0.2007$. 6. So $\frac{x}{3} = 0.2838 \pm 0.2007 + 2k\pi$. 7. For $k=0$, $x = 3(0.4845) = 1.4535$ or $x = 3(0.0831) = 0.2493$. 8. Check $0 < x < \pi \approx 3.1416$, both valid. ### Problem 3 1. Given $\tan (x - 60^\circ) = 2 \cot x$. 2. Use $\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$. 3. Let $t = \tan x$, then $\tan (x - 60^\circ) = \frac{t - \sqrt{3}}{1 + \sqrt{3} t}$. 4. Also, $\cot x = \frac{1}{t}$. 5. Equation: $\frac{t - \sqrt{3}}{1 + \sqrt{3} t} = \frac{2}{t}$. 6. Cross multiply: $t (t - \sqrt{3}) = 2 (1 + \sqrt{3} t)$. 7. Expand: $t^2 - \sqrt{3} t = 2 + 2 \sqrt{3} t$. 8. Rearrange: $t^2 - \sqrt{3} t - 2 - 2 \sqrt{3} t = 0$. 9. Combine terms: $t^2 - 3 \sqrt{3} t - 2 = 0$. 10. Solve quadratic: $t = \frac{3 \sqrt{3} \pm \sqrt{(3 \sqrt{3})^2 + 8}}{2}$. 11. Calculate discriminant: $27 + 8 = 35$. 12. So $t = \frac{3 \sqrt{3} \pm \sqrt{35}}{2}$. 13. Find $x = \arctan t$ in $[0^\circ, 180^\circ]$. ### Problem 4 (a) 1. Given $\tan^3 x + 2 \tan 2x - \tan x = 0$. 2. Use $\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$. 3. Substitute: $\tan^3 x + 2 \cdot \frac{2 \tan x}{1 - \tan^2 x} - \tan x = 0$. 4. Multiply both sides by $1 - \tan^2 x$: $\tan^3 x (1 - \tan^2 x) + 4 \tan x - \tan x (1 - \tan^2 x) = 0$. 5. Expand: $\tan^3 x - \tan^5 x + 4 \tan x - \tan x + \tan^3 x = 0$. 6. Simplify: $- \tan^5 x + 2 \tan^3 x + 3 \tan x = 0$. 7. Divide by $\tan x \neq 0$: $- \tan^4 x + 2 \tan^2 x + 3 = 0$. 8. Multiply by $-1$: $\tan^4 x - 2 \tan^2 x - 3 = 0$. (b) 1. Given $\tan^3 2\theta + 2 \tan 4\theta - \tan 2\theta = 0$. 2. Replace $x$ by $2\theta$ in part (a) result: $\tan^4 2\theta - 2 \tan^2 2\theta - 3 = 0$. 3. Let $t = \tan^2 2\theta$, then $t^2 - 2t - 3 = 0$. 4. Solve: $t = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm 4}{2}$. 5. So $t = 3$ or $t = -1$ (discard negative). 6. $\tan^2 2\theta = 3 \Rightarrow \tan 2\theta = \pm \sqrt{3}$. 7. Solve $2\theta = 60^\circ, 120^\circ, 240^\circ, 300^\circ$. 8. So $\theta = 30^\circ, 60^\circ, 120^\circ, 150^\circ$ in $(0, \pi)$. ### Problem 5 (a) 1. Express $3 \sin x + 2 \sqrt{2} \cos \left(x + \frac{\pi}{4}\right)$ as $R \sin (x + \alpha)$. 2. Expand $\cos \left(x + \frac{\pi}{4}\right) = \cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} (\cos x - \sin x)$. 3. Substitute: $3 \sin x + 2 \sqrt{2} \cdot \frac{\sqrt{2}}{2} (\cos x - \sin x) = 3 \sin x + 2 (\cos x - \sin x) = (3 - 2) \sin x + 2 \cos x = \sin x + 2 \cos x$. 4. So expression is $\sin x + 2 \cos x$. 5. Find $R = \sqrt{1^2 + 2^2} = \sqrt{5}$. 6. Find $\alpha = \arctan \frac{1}{2} \approx 0.464$ radians (3 decimals). (b) 1. Solve $6 \sin \frac{\theta}{2} + 4 \sqrt{2} \cos \left( \frac{\theta}{2} + \frac{\pi}{4} \right) = 3$. 2. Use same expansion as (a): $4 \sqrt{2} \cos \left( \frac{\theta}{2} + \frac{\pi}{4} \right) = 4 \sqrt{2} \cdot \frac{\sqrt{2}}{2} (\cos \frac{\theta}{2} - \sin \frac{\theta}{2}) = 4 (\cos \frac{\theta}{2} - \sin \frac{\theta}{2})$. 3. Equation becomes $6 \sin \frac{\theta}{2} + 4 \cos \frac{\theta}{2} - 4 \sin \frac{\theta}{2} = 3$. 4. Simplify: $(6 - 4) \sin \frac{\theta}{2} + 4 \cos \frac{\theta}{2} = 3$. 5. So $2 \sin \frac{\theta}{2} + 4 \cos \frac{\theta}{2} = 3$. 6. Express as $R \sin \left( \frac{\theta}{2} + \alpha \right)$ with $R = \sqrt{2^2 + 4^2} = \sqrt{20} = 2 \sqrt{5}$. 7. $\alpha = \arctan \frac{4}{2} = \arctan 2 \approx 1.107$ radians. 8. Equation: $2 \sqrt{5} \sin \left( \frac{\theta}{2} + 1.107 \right) = 3$. 9. Divide: $\sin \left( \frac{\theta}{2} + 1.107 \right) = \frac{3}{2 \sqrt{5}} = 0.6708$. 10. Solve: $\frac{\theta}{2} + 1.107 = \arcsin 0.6708 + 2k\pi$ or $\pi - \arcsin 0.6708 + 2k\pi$. 11. $\arcsin 0.6708 \approx 0.735$. 12. So $\frac{\theta}{2} = -1.107 + 0.735 + 2k\pi = -0.372 + 2k\pi$ or $\frac{\theta}{2} = -1.107 + (\pi - 0.735) + 2k\pi = 1.299 + 2k\pi$. 13. Multiply by 2: $\theta = -0.744 + 4k\pi$ or $2.598 + 4k\pi$. 14. For $-4\pi < \theta < 4\pi$, list all $k$ values and solutions. Final answers: 1. $x = 26.565^\circ, 45^\circ$. 2. (a) $R=25$, $\alpha=0.2838$ radians. (b) $x \approx 0.2493, 1.4535$. 3. $\tan x = \frac{3 \sqrt{3} \pm \sqrt{35}}{2}$, find $x$ in $[0^\circ,180^\circ]$. 4. (a) $\tan^4 x - 2 \tan^2 x - 3 = 0$. (b) $\theta = 30^\circ, 60^\circ, 120^\circ, 150^\circ$. 5. (a) $R=\sqrt{5}$, $\alpha=0.464$ radians. (b) $\theta = -0.744 + 4k\pi$ or $2.598 + 4k\pi$ for integers $k$ with $-4\pi < \theta < 4\pi$.