1. We are given the curve $y = \cos x \sin 2x$ and need to find the stationary points where $0 < x < \frac{1}{2} \pi$. A stationary point occurs where $\frac{dy}{dx} = 0$. 2. Differentiate $y$ using the product rule: $$y = \cos x \sin 2x$$ $$\frac{dy}{dx} = \frac{d}{dx}(\cos x) \sin 2x + \cos x \frac{d}{dx}(\sin 2x)$$ $$= (-\sin x) \sin 2x + \cos x (2 \cos 2x)$$ 3. Set $\frac{dy}{dx} = 0$: $$-\sin x \sin 2x + 2 \cos x \cos 2x = 0$$ 4. Use the double angle identity $\sin 2x = 2 \sin x \cos x$: $$-\sin x (2 \sin x \cos x) + 2 \cos x \cos 2x = 0$$ $$-2 \sin^2 x \cos x + 2 \cos x \cos 2x = 0$$ 5. Factor out $2 \cos x$: $$2 \cos x (-\sin^2 x + \cos 2x) = 0$$ 6. Since $0 < x < \frac{1}{2} \pi$, $\cos x \neq 0$, so: $$-\sin^2 x + \cos 2x = 0$$ 7. Use identity $\cos 2x = 1 - 2 \sin^2 x$: $$-\sin^2 x + 1 - 2 \sin^2 x = 0$$ $$1 - 3 \sin^2 x = 0$$ $$\sin^2 x = \frac{1}{3}$$ 8. Solve for $x$: $$\sin x = \pm \frac{1}{\sqrt{3}}$$ Within $0 < x < \frac{\pi}{2}$, only positive root applies: $$x = \arcsin \left( \frac{1}{\sqrt{3}} \right) \approx 0.615$$ --- 9. For the second problem, $y = e^{-5x} \tan^2 x$ with $-\frac{\pi}{2} < x < \frac{\pi}{2}$. 10. Differentiate using product and chain rules: $$y = e^{-5x} (\tan x)^2$$ $$\frac{dy}{dx} = e^{-5x} \cdot 2 \tan x \sec^2 x + (\tan^2 x)(-5 e^{-5x})$$ $$= e^{-5x} (2 \tan x \sec^2 x - 5 \tan^2 x)$$ 11. Set derivative to zero: $$e^{-5x} (2 \tan x \sec^2 x - 5 \tan^2 x) = 0$$ Since $e^{-5x} \neq 0$: $$2 \tan x \sec^2 x - 5 \tan^2 x = 0$$ 12. Factor $\tan x$: $$\tan x (2 \sec^2 x - 5 \tan x) = 0$$ 13. Solutions: - $\tan x = 0 \Rightarrow x = 0$ - $2 \sec^2 x - 5 \tan x = 0$ 14. Use $\sec^2 x = 1 + \tan^2 x$: $$2 (1 + \tan^2 x) - 5 \tan x = 0$$ $$2 + 2 \tan^2 x - 5 \tan x = 0$$ 15. Let $t = \tan x$: $$2 t^2 - 5 t + 2 = 0$$ 16. Solve quadratic: $$t = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4}$$ 17. Roots: $$t = 2, \quad t = \frac{1}{2}$$ 18. Find $x$: $$x = \arctan 0 = 0$$ $$x = \arctan 2 \approx 1.107$$ $$x = \arctan \frac{1}{2} \approx 0.464$$ All within $-\frac{\pi}{2} < x < \frac{\pi}{2}$. --- 19. For the third problem, $y = \frac{e^{\sin x}}{\cos^2 x}$, $0 \leq x \leq 2\pi$. 20. Rewrite: $$y = e^{\sin x} \sec^2 x$$ 21. Differentiate using product rule: $$\frac{dy}{dx} = e^{\sin x} \cos x \sec^2 x + e^{\sin x} \cdot 2 \sec^2 x \tan x$$ 22. Simplify: $$\frac{dy}{dx} = e^{\sin x} \sec^2 x (\cos x + 2 \tan x)$$ 23. Set derivative to zero: $$e^{\sin x} \sec^2 x (\cos x + 2 \tan x) = 0$$ Since $e^{\sin x} \neq 0$ and $\sec^2 x \neq 0$ where defined, 24. Solve: $$\cos x + 2 \tan x = 0$$ 25. Use $\tan x = \frac{\sin x}{\cos x}$: $$\cos x + 2 \frac{\sin x}{\cos x} = 0$$ $$\cos^2 x + 2 \sin x = 0$$ 26. Use $\cos^2 x = 1 - \sin^2 x$: $$1 - \sin^2 x + 2 \sin x = 0$$ 27. Let $s = \sin x$: $$-s^2 + 2 s + 1 = 0$$ Multiply by -1: $$s^2 - 2 s - 1 = 0$$ 28. Solve quadratic: $$s = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$$ 29. Since $\sin x$ must be between -1 and 1, only $s = 1 - \sqrt{2} \approx -0.414$ is valid. 30. Find $x$: $$x = \arcsin(-0.414) \approx -0.427$$ But $x$ must be in $[0, 2\pi]$, so consider the second solution in the range: $$x = \pi - (-0.427) = 3.569$$ 31. Check domain for $\sec^2 x$ (no division by zero): $\cos x \neq 0$, so exclude $x = \frac{\pi}{2}, \frac{3\pi}{2}$. Final answers: 1. $x \approx 0.615$ 2. $x = 0, 0.464, 1.107$ 3. $x \approx 3.569$