The equation $\sqrt 3 \sin x + \cos x = 4$ has

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

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

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

The number of solutions of $|\cos x|=\sin x$, such that $-4 \pi \leq x \leq 4 \pi$ is.

Let $f(x) = sinx + 2sin^2x + 3sin^3x + 4sin^4x+....\infty $ , then number of solution $(s)$ of equation $f(x) = 2$ in $x \in \left[ { - \pi ,\pi } \right] - \left\{ { \pm \frac{\pi }{2}} \right\}$ is

The variable $x$ satisfying the equation $\left| {\sin \,x\,\cos \,x} \right| + \sqrt {2 + {{\tan }^2}\,x + {{\cot }^2}\,x}  = \sqrt 3$ belongs to the interval

If $(1 + \tan \theta )(1 + \tan \phi ) = 2$, then $\theta + \phi  =$ ....$^o$