The sum of solutions in $x \in (0,2\pi )$ of the equation, $4\cos (x).\cos \left( {\frac{\pi }{3} - x} \right).\cos \left( {\frac{\pi }{3} + x} \right) = 1$ is equal to
$\pi $
$2\pi $
$3\pi $
$4\pi $
The sides of a triangle are $\sin \alpha ,\,\cos \alpha $ and $\sqrt {1 + \sin \alpha \cos \alpha } $ for some $0 < \alpha < \frac{\pi }{2}$. Then the greatest angle of the triangle is.....$^o$
The number of solutions of $|\cos x|=\sin x$, such that $-4 \pi \leq x \leq 4 \pi$ is.
The number of values of $\theta$ in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\theta \neq \frac{n \pi}{5}$ for $n=0, \pm 1, \pm 2$ and $\tan \theta=\cot 5 \theta$ as well as $\sin 2 \theta=\cos 4 \theta$ is
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
Number of solutions of $8cosx$ = $x$ will be