Question 9:

Find the fraction of the interval $0 \leq \theta \leq 2\pi$ for which the inequality

$$(\cos\theta + \sin\theta) \left(\tfrac{\sqrt{3}}{2} + \cos\tfrac{1}{2}\theta\right) (\cos^{2}\theta - 1) \leq 0$$

Solution:

1. Rewrite: $\cos\theta + \sin\theta = \sqrt{2}\sin\!\left(\theta + \tfrac{\pi}{4}\right)$.
2. This term is negative for $\tfrac{3\pi}{4} < \theta < \tfrac{7\pi}{4}$.
3. Solve $\tfrac{\sqrt{3}}{2} + \cos\tfrac{1}{2}\theta = 0$. This gives $\tfrac{1}{2}\theta = \tfrac{5\pi}{6} \implies \theta = \tfrac{5\pi}{3}$.
4. Finally, $\cos^{2}\theta - 1 \leq 0$ for all $\theta$.

From sign analysis on intervals $0 \leq \theta \leq 2\pi$, the inequality holds for: $$0 \leq \theta \leq \tfrac{3\pi}{4}, \quad \tfrac{5\pi}{3} \leq \theta \leq \tfrac{7\pi}{4}.$$

This totals a length of: $$\tfrac{3\pi}{4} + \left(\tfrac{7\pi}{4} - \tfrac{5\pi}{3}\right) = \tfrac{5\pi}{6}.$$

Fraction of interval: $$\frac{\tfrac{5\pi}{6}}{2\pi} = \tfrac{5}{12}.$$

Answer: (B) $\tfrac{5}{12}$