Question

Sol. [Key Concept: Simplify $2(\cos^{8}\theta - \sin^{8}\theta)sec\ 2\theta$; it cannot equal an integer square $\Rightarrow \ n(S) = 0$.]

$$2\left( \cos^{8}\theta - \sin^{8}\theta \right)sec2\theta = a^{2}$$

$$\begin{matrix} & 2\left( \cos^{4}\theta + \sin^{4}\theta \right)\left( \cos^{2}\theta + \sin^{2}\theta \right)\left( \cos^{2}\theta - \sin^{2}\theta \right)sec2\theta = a^{2} \\ & 2\left( \cos^{4}\theta + \sin^{4}\theta \right) = a^{2} \\ & 2\left( 1 - 2\sin^{2}\theta\cos^{2}\theta \right) = a^{2} \\ & 2\left( 1 - \frac{\sin^{2}2\theta}{2} \right) = a^{2},a \in Z \\ & a^{2} = 2 - \sin^{2}2\theta \in \lbrack 1,2\rbrack \\ & a^{2} = 1\text{~}\text{at}\text{~}\sin^{2}2\theta = 1 \\ & 2\theta = (2n + 1)\frac{\pi}{2} \\ & \theta = (2n + 1)\frac{\pi}{4} \notin P \\ & n(\text{ }S) = 0 \end{matrix}$$