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 $|\cos x + \sin x| + |\cos x - \sin x| = 2 \sin x$ for $x \in [0, 2\pi]$,then the maximum integral value of $x$ is:

If $y^2+z^2=3yz$,$z^2+x^2=8zx$,and $x^2+y^2=4xy$,then the value of $\frac{y^2}{xz}+\frac{xz}{y^2}$ is

If $\theta$ is eliminated from the equations $x = a \cos(\theta - \alpha)$ and $y = b \cos(\theta - \beta)$,then $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos(\alpha - \beta)$ is equal to

If $A$ and $B$ are positive acute angles satisfying $3 \cos^2 A + 2 \cos^2 B = 4$ and $\frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}$,then $A + 2B =$ (in $^{\circ}$)

Evaluate the product: $\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{4 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{6 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right)$