Solve $2 \cos ^{2} x+3 \sin x=0$
The equation can be written as
$2\left(1-\sin ^{2} x\right)+3 \sin x=0$
or $2 \sin ^{2} x-3 \sin x-2=0$
or $(2 \sin x+1)(\sin x-2)=0$
Hence $\sin x=-\frac{1}{2} \quad$ or $\quad \sin x=2$
But $\sin x=2$ is not possible (Why?)
Therefore $\sin x=-\frac{1}{2}=\sin \frac{7 \pi}{6}$
Hence, the solution is given by
$x=n \pi+(-1)^{n} \frac{7 \pi}{6}, \text { where } n \in Z.$
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
$\cot \theta = \sin 2\theta (\theta \ne n\pi $, $n$ is integer), if $\theta = $
If $\sec x\cos 5x + 1 = 0$, where $0 < x < 2\pi $, then $x =$
Number of solution$(s)$ of the equation $\sin 2\theta + \cos 2\theta = - \frac{1}{2},\theta \in \left( {0,\frac{\pi }{2}} \right)$ is-
If $2{\tan ^2}\theta = {\sec ^2}\theta ,$ then the general value of $\theta $ is