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

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

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

If $\theta $ and $\phi $ are acute satisfying $\sin \theta = \frac{1}{2},$ $\cos \phi = \frac{1}{3},$ then $\theta + \phi \in $

The roots of the equation $1 - \cos \theta = \sin \theta .\sin \frac{\theta }{2}$ is

The number of points in $(-\infty, \infty)$, for which $x^2-x \sin x-\cos x=0$, is

The solution of the equation ${\cos ^2}x - 2\cos x = $ $4\sin x - \sin 2x,$ $\,(0 \le x \le \pi )$ is

The number of solutions of the equation $1 + {\sin ^4}\,x = {\cos ^2}\,3x,x\,\in \,\left[ { - \frac{{5\pi }}{2},\frac{{5\pi }}{2}} \right]$ is