If ${\sin ^2}\theta = \frac{1}{4},$ then the most general value of $\theta $ is
$2n\pi \pm {( - 1)^n}\frac{\pi }{6}$
$\frac{{n\pi }}{2} \pm {( - 1)^n}\frac{\pi }{6}$
$n\pi \pm \frac{\pi }{6}$
$2n\pi \pm \frac{\pi }{6}$
The solution of the equation ${\cos ^2}x - 2\cos x = $ $4\sin x - \sin 2x,$ $\,(0 \le x \le \pi )$ is
Let $A=\left\{\theta \in R:\left(\frac{1}{3} \sin \theta+\frac{2}{3} \cos \theta\right)^2=\frac{1}{3} \sin ^2 \theta+\frac{2}{3} \cos ^2 \theta\right\}$.Then
If $\sin 3\alpha = 4\sin \alpha \sin (x + \alpha )\sin (x - \alpha ),$ then $x = $
Let $\theta \in [0, 4\pi ]$ satisfy the equation $(sin\, \theta + 2) (sin\, \theta + 3) (sin\, \theta + 4) = 6$ . If the sum of all the values of $\theta $ is of the form $k\pi $, then the value of $k$ is
If $2(\sin x - \cos 2x) - \sin 2x(1 + 2\sin x)2\cos x = 0$ then