If $2(\sin x - \cos 2x) - \sin 2x(1 + 2\sin x)2\cos x = 0$ then
$x = \frac{\pi }{6}(4n + 1)$or $x = \frac{\pi }{2}(4n - 1)$
$x = \frac{\pi }{6}(4n - 1)$or $x = \frac{\pi }{2}(4n - 1)$
$x = \frac{\pi }{6}(4n + 1)$or $x = \frac{\pi }{2}(4n + 1)$
None of these
If $\cot \theta + \tan \theta = 2{\rm{cosec}}\theta $, the general value of $\theta $ is
The numbers of solution $(s)$ of the equation $\left( {1 - \frac{1}{{2\,\sin x}}} \right){\cos ^2}\,2x\, = \,2\,\sin x\, - \,3\, + \,\frac{1}{{\sin x}}$ in $[0,4\pi ]$ is
The number of solutions of the equation $2 \theta-\cos ^{2} \theta+\sqrt{2}=0$ is $R$ is equal to
The angles $\alpha, \beta, \gamma$ of a triangle satisfy the equations $2 \sin \alpha+3 \cos \beta=3 \sqrt{2}$ and $3 \sin \beta+2 \cos \alpha=1$. Then, angle $\gamma$ equals
If $\cos p\theta = \cos q\theta ,p \ne q$, then