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Let $|\cos \theta \cos (60^{\circ}-\theta) \cos (60^{\circ}+\theta)| \leq \frac{1}{8}$, where $\theta \in [0, 2\pi]$. Then, the sum of all $\theta \in [0, 2\pi]$ where $\cos 3\theta$ attains its maximum value is: (in $\pi$)

Let $S$ be the set of all $\alpha \in \mathbb{R}$ such that the equation $\cos 2x + \alpha \sin x = 2\alpha - 7$ has a solution. Then $S$ is equal to

If $\pi < \alpha < \frac{3\pi}{2}$,then $\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = $

If $A(n) = \sin^n \alpha + \cos^n \alpha$,then $A(1) A(4) + A(2) A(5) =$

$\frac{1}{\sin 1^{\circ} \sin 2^{\circ}}+\frac{1}{\sin 2^{\circ} \sin 3^{\circ}}+\frac{1}{\sin 3^{\circ} \sin 4^{\circ}}+\ldots+\frac{1}{\sin 89^{\circ} \sin 90^{\circ}} = ?$

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