The number of solution of the equation,$\sum\limits_{r = 1}^5 {\cos (r\,x)} $ $= 0$ lying in $(0, \pi)$ is :

If $\sqrt 3 \cos \,\theta + \sin \theta = \sqrt 2 ,$ then the most general value of $\theta $ is

Let $P = \left\{ {\theta :\sin \,\theta  - \cos \,\theta  = \sqrt 2 \,\cos \,\theta } \right\}$ and $Q = \left\{ {\theta :\sin \,\theta  + \cos \,\theta  = \sqrt {2\,} \sin \,\theta } \right\}$ be two sets. Then

If $|k|\, = 5$ and ${0^o} \le \theta \le {360^o}$, then the number of different solutions of $3\cos \theta + 4\sin \theta = k$ is

The number of values of $x$ in the interval $[0, 5\pi]$ satisfying the equation $3sin^2x\, \,-\,\, 7sinx + 2 = 0$ is

If $5{\cos ^2}\theta + 7{\sin ^2}\theta - 6 = 0$, then the general value of $\theta $ is