The number of elements in the set $S=\left\{x \in R : 2 \cos \left(\frac{x^{2}+x}{6}\right)=4^{x}+4^{-x}\right\}$ is$…..$

If $4{\sin ^2}\theta + 2(\sqrt 3 + 1)\cos \theta = 4 + \sqrt 3 $, then the general value of $\theta $ is

The total number of solution of $sin^4x + cos^4x = sinx\, cosx$ in $[0, 2\pi ]$ is equal to

If $\cos \,x = \frac{{2\cos y – 1}}{{2 – \cos y}},x,\,y\, \in \,\left( {0,\pi } \right),$ then $tan(x/2)cot(y/2) =$

Let $\theta, \phi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \phi)=\sin ^2 \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \phi-1, \tan (2 \pi-\theta)>0$ and $-1 < \sin \theta < -\frac{\sqrt{3}}{2}$. Then $\phi$ cannot satisfy

$(A)$ $0 < \phi<\frac{\pi}{2}$ $(B)$ $\frac{\pi}{2} < \phi<\frac{4 \pi}{3}$

$(C)$ $\frac{4 \pi}{3} < \phi<\frac{3 \pi}{2}$ $(D)$ $\frac{3 \pi}{2} < \phi < 2 \pi$