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 $\cos 3x + \sin \left( {2x - \frac{{7\pi }}{6}} \right) = - 2$, then $x = $ (where $k \in Z$)
The total number of solution of $sin^4x + cos^4x = sinx\, cosx$ in $[0, 2\pi ]$ is equal to
If $L=\sin ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$ and $M=\cos ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right),$ then
The number of solutions to the equation $\cos ^4 x+\frac{1}{\cos ^2 x}=\sin ^4 x+\frac{1}{\sin ^2 x}$ in the interval $[0,2 \pi]$ is
The general value of $\theta $ satisfying ${\sin ^2}\theta + \sin \theta = 2$ is