Question

Solve:$2co{s}^{2}x+3sinx=0$

Answer 1

We have

$2co{s}^{2}x+3sinx=0$

$\Rightarrow 2(1-si{n}^{2}x)+3sinx=0$

$\Rightarrow 2si{n}^{2}x-3sinx-2=0$

$\Rightarrow 2si{n}^{2}x-4sinx+sinx-2=0$

$\Rightarrow 2sinx(sinx-2)+(sinx-2)=0$

$\Rightarrow (sinx-2)(2sinx+1)=0$

$\Rightarrow (sinx-2)=0$ or $(2sinx+1)=0$

$\Rightarrow sinx=2$ or $(2sinx+1)=0$

$\Rightarrow 2sinx+1=0[\because sinx=2$ is not possible]

$\Rightarrow sinx=-\frac{1}{2}=-sin\frac{\pi }{6}=sin(\pi +\frac{\pi }{6})=sin\frac{7\pi }{6}$

$\Rightarrow sinx=sin\frac{7\pi }{6}$

$\Rightarrow x=\{n\pi +(-1{)}^n.\frac{7\pi }{6}\}$, where $n\in I.$

Hence, the general solution is given by $x=\{n\pi +(-1{)}^n.\frac{7\pi }{6}\}$, where $n\in I$.