Find the principal solutions of the equation $\sin x=\frac{\sqrt{3}}{2}$
We know that, $\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$ and $\sin \frac{2 \pi}{3}=\sin \left(\pi-\frac{\pi}{3}\right)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$
Therefore, principal solutions are $x=\frac{\pi}{3}$ and $\frac{2 \pi}{3}$.
In a triangle $P Q R, P$ is the largest angle and $\cos P=\frac{1}{3}$. Further the incircle of the triangle touches the sides $P Q, Q R$ and $R P$ at $N, L$ and $M$ respectively, such that the lengths of $P N, Q L$ and $R M$ are consecutive even integers. Then possible length$(s)$ of the side$(s)$ of the triangle is (are)
$(A)$ $16$ $(B)$ $18$ $(C)$ $24$ $(D)$ $22$
If $\sin 3\alpha = 4\sin \alpha \sin (x + \alpha )\sin (x - \alpha ),$ then $x = $
The values of $\theta $ satisfying $\sin 7\theta = \sin 4\theta - \sin \theta $ and $0 < \theta < \frac{\pi }{2}$ are
If $\tan \theta + \tan 2\theta + \sqrt 3 \tan \theta \tan 2\theta = \sqrt 3 ,$ then
If $3({\sec ^2}\theta + {\tan ^2}\theta ) = 5$, then the general value of $\theta $ is