$\cos^2\theta(\cos^2\theta - 3\cos\theta + 2) = 1$, if $\theta$ belongs to
Step-by-Step Solution
Key Concept: Manipulate the inequality into a factorable form and analyze the sign of each factor using properties of cosine.
Starting with $\cos^6\theta - 3\cos\theta + 2 \geq \frac{1}{\cos^2\theta}$ and $\sin^2\theta$. Rearranging: $\cos^6\theta - 3\cos\theta + 2 \geq 1 - \cos^2\theta$, which simplifies to $2\cos^6\theta - 3\cos\theta + 1 \geq 0$ or $(2\cos\theta - 1)(\cos^2\theta - 1) \geq 0$. Since $\cos^2\theta \leq 1$, we need $\cos\theta \leq \frac{1}{2}$. For $\theta \in (\frac{7\pi}{12}, \frac{7\pi}{12})$ of given intervals, the solution set is $\theta \in (\frac{7\pi}{12}, \frac{7\pi}{12})$.
Correct Answer: 2