Question

Solve $2{\cos }^{2}x+3\sin x=0$

Answer 1

Step 1 : Simplification
$2{\cos }^{2}x+3\sin x=0$
$\Rightarrow 2(1-{\sin }^{2}x)+3\sin x=0$
$\Rightarrow 2-2{\sin }^{2}x+3\sin x=0$
$\Rightarrow 2{\sin }^{2}x-3\sin x-2=0$
$\Rightarrow 2{\sin }^{2}x-4\sin x+\sin x-2=0$
$\Rightarrow 2\sin x(\sin x-2)+1(\sin x-2),=0$
$\Rightarrow (2\sin x+1)(\sin x-2)=0$
$\Rightarrow 2\sin x+1=0$ or $\sin x-2=0$
$\Rightarrow \sin x=-\frac{1}{2}$ or $\sin x=2\text{(not possible)}$
$\therefore \sin x=-\frac{1}{2}$

Step 2:
$\sin x=-\frac{1}{2}$
$\sin x=-\sin \frac{\pi }{6}=\sin \left(\frac{7\pi }{6}\right)$
$\Rightarrow x=n\pi +(-1{)}^n\left(\frac{7\pi }{6}\right)$ where $n\in Z$

General solution of $2{\cos }^{2}x+3\sin x=0$ is $x=n\pi +(-1{)}^n\left(\frac{7\pi }{6}\right)$ where $n\in Z$