The solution of the equation $4{\cos ^2}x + 6$${\sin ^2}x = 5$

If $\sqrt{3}\left(\cos ^{2} x\right)=(\sqrt{3}-1) \cos x+1,$ the number of solutions of the given equation when $x \in\left[0, \frac{\pi}{2}\right]$ is

Let $S=\left\{\theta \in(0,2 \pi): 7 \cos ^{2} \theta-3 \sin ^{2} \theta-2\right.$ $\left.\cos ^{2} 2 \theta=2\right\}$. Then, the sum of roots of all the equations $x ^{2}-2\left(\tan ^{2} \theta+\cot ^{2} \theta\right) x +6 \sin ^{2} \theta=0$ $\theta \in S$, is$...$

Number of solutions of $\sqrt {\tan \theta }  = 2\sin \theta ,\theta  \in \left[ {0,2\pi } \right]$ is equal to 

If $\frac{{1 - \cos 2\theta }}{{1 + \cos 2\theta }} = 3$, then the general value of $\theta $ is

Let $\theta, 0 < \theta < \pi / 2$, be an angle such that the equation $x ^2+4 x \cos \theta+\cot \theta=0$ has equal roots for $x$. Then $\theta$ in radians is