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 $A, B, C, D$ are the angles of a cyclic quadrilateral taken in order, then
$cos(180^o + A) + cos(180^o -B) + cos(180^o -C) -sin(90^o -D)=$
The number of solutions of the equation $\cos \left(x+\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x, x \in[-3 \pi$ $3 \pi]$ is
If ${\left( {\frac{{\sin \theta }}{{\sin \phi }}} \right)^2} = \frac{{\tan \theta }}{{\tan \phi }} = 3,$ then the value of $\theta $ and $\phi $ are
If $\sin \theta + \cos \theta = 1$ then the general value of $\theta $ is
All possible values of $\theta \in[0,2 \pi]$ for which $\sin 2 \theta+\tan 2 \theta>0$ lie in