1. **Problem (a): Solve $\cos 2x + \cos x = 0$.** Use the double-angle formula: $\cos 2x = 2\cos^2 x - 1$. Rewrite the equation: $$2\cos^2 x - 1 + \cos x = 0$$ Let $y = \cos x$, then: $$2y^2 + y - 1 = 0$$ Solve quadratic: $$y = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$$ So, $$y_1 = \frac{2}{4} = \frac{1}{2}, \quad y_2 = \frac{-4}{4} = -1$$ Find $x$: - For $\cos x = \frac{1}{2}$, solutions are: $$x = \pm \frac{\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}$$ - For $\cos x = -1$, solution is: $$x = \pi + 2k\pi, \quad k \in \mathbb{Z}$$ --- 2. **Problem (b): Given $\cos \theta = \frac{5}{13}$, $\frac{3\pi}{2} < \theta < 2\pi$, find $\sin 2\theta$, $\cos 2\theta$, $\tan 2\theta$.** Since $\theta$ is in the fourth quadrant, $\sin \theta < 0$. Find $\sin \theta$ using Pythagoras: $$\sin \theta = -\sqrt{1 - \cos^2 \theta} = -\sqrt{1 - \left(\frac{5}{13}\right)^2} = -\sqrt{1 - \frac{25}{169}} = -\sqrt{\frac{144}{169}} = -\frac{12}{13}$$ Use double-angle formulas: $$\sin 2\theta = 2 \sin \theta \cos \theta = 2 \times \left(-\frac{12}{13}\right) \times \frac{5}{13} = -\frac{120}{169}$$ $$\cos 2\theta = \cos^2 \theta - \sin^2 \theta = \left(\frac{5}{13}\right)^2 - \left(-\frac{12}{13}\right)^2 = \frac{25}{169} - \frac{144}{169} = -\frac{119}{169}$$ $$\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{-\frac{120}{169}}{-\frac{119}{169}} = \frac{120}{119}$$ --- 3. **Problem (c): Given $\sin \theta = \frac{3}{5}$, $0 < \theta < \frac{\pi}{2}$, find $\sin 2\theta$, $\cos 2\theta$, $\tan 2\theta$.** Since $\theta$ is in the first quadrant, $\cos \theta > 0$. Find $\cos \theta$: $$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$ Use double-angle formulas: $$\sin 2\theta = 2 \sin \theta \cos \theta = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}$$ $$\cos 2\theta = \cos^2 \theta - \sin^2 \theta = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$$ $$\tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{\frac{24}{25}}{\frac{7}{25}} = \frac{24}{7}$$ --- 4. **Problem (d): Rewrite $\cos 3x$ in terms of $\cos x$.** Use triple-angle formula: $$\cos 3x = 4 \cos^3 x - 3 \cos x$$ --- 5. **Problem (e): Rewrite $\tan^4 x$ in terms of first powers of cosines of multiple angles.** Start with: $$\tan^4 x = \left(\frac{\sin x}{\cos x}\right)^4 = \frac{\sin^4 x}{\cos^4 x}$$ Use power-reduction formulas: $$\sin^4 x = \left(\sin^2 x\right)^2 = \left(\frac{1 - \cos 2x}{2}\right)^2 = \frac{1 - 2 \cos 2x + \cos^2 2x}{4}$$ Similarly, $$\cos^4 x = \left(\cos^2 x\right)^2 = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1 + 2 \cos 2x + \cos^2 2x}{4}$$ Rewrite $\cos^2 2x$: $$\cos^2 2x = \frac{1 + \cos 4x}{2}$$ Substitute back: $$\sin^4 x = \frac{1 - 2 \cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{2 - 4 \cos 2x + 1 + \cos 4x}{8} = \frac{3 - 4 \cos 2x + \cos 4x}{8}$$ $$\cos^4 x = \frac{1 + 2 \cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{2 + 4 \cos 2x + 1 + \cos 4x}{8} = \frac{3 + 4 \cos 2x + \cos 4x}{8}$$ Therefore: $$\tan^4 x = \frac{\sin^4 x}{\cos^4 x} = \frac{3 - 4 \cos 2x + \cos 4x}{3 + 4 \cos 2x + \cos 4x}$$ This expresses $\tan^4 x$ in terms of first powers of cosines of multiple angles. --- **Final answers:** (a) $x = \pm \frac{\pi}{3} + 2k\pi$ or $x = \pi + 2k\pi$ (b) $\sin 2\theta = -\frac{120}{169}$, $\cos 2\theta = -\frac{119}{169}$, $\tan 2\theta = \frac{120}{119}$ (c) $\sin 2\theta = \frac{24}{25}$, $\cos 2\theta = \frac{7}{25}$, $\tan 2\theta = \frac{24}{7}$ (d) $\cos 3x = 4 \cos^3 x - 3 \cos x$ (e) $\tan^4 x = \frac{3 - 4 \cos 2x + \cos 4x}{3 + 4 \cos 2x + \cos 4x}$