Solve $\cos x=\frac{1}{2}$

If $K = sin^6x + cos^6x$, then $K$ belongs to the interval

The solution set of $(5 + 4\cos \theta )(2\cos \theta + 1) = 0$ in the interval $[0,\,\,2\pi ]$ is

If sum of all the solutions of the equation $8\cos x \cdot \left( {\cos \left( {\frac{\pi }{6} + x} \right) \cdot \cos \left( {\frac{\pi }{6} - x} \right) - \frac{1}{2}} \right) = 1$ in $\left[ {0,\pi } \right]$ is $k\pi $then $k$ is equal to :

If $\sin (A + B) =1 $ and $\cos (A - B) = \frac{{\sqrt 3 }}{2},$ then the smallest positive values of $A$ and $ B$ are

Let $S=\left\{\theta \in[-\pi, \pi]-\left\{\pm \frac{\pi}{2}\right\}: \sin \theta \tan \theta+\tan \theta=\sin 2 \theta\right\} \text {. }$  If $T =\sum_{\theta \in S } \cos 2 \theta$, then $T + n ( S )$ is equal