Question

Find the general solution of the following questions
$\cos x=2x$
$\sin 2x+\cos x=0$.

Answer 1

$\sin 2x+cosx=0$

$\because \sin 2x=2sinxcosx$

$\Rightarrow 2sinxcosx+cosx=0$

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

$\swarrow$$\searrow$

$cosx=0$ $2sinx=-1$

$sinx=-1/2$

General solution for $cosx=0$

$x=(2n+1)\pi /2,n\in z$

$\Rightarrow x=2n\pi \pm \pi /2,n\in z$

General solution for $sinx=-1/2$

let $sinx=siny\to \left(1\right)$

$sinx=-1/2\to \left(2\right)$

From $\left(1\right)$ & $\left(2\right)$

$siny=-1/2$

$siny=\sin \frac{7\pi }{6}$

$y=\frac{7\pi }{6}$

General solution $=x=n\pi +{(-1)}^{n}y,n\in z$

put $y=\frac{7\pi }{6}$

$x=n\pi +{(-1)}^n\frac{7\pi }{6},n\in z$

Hence for $cosx=0,x=(2n+1)\pi /2,n\in z$

for $sinx=-1/2,x=n\pi +{(-1)}^n\frac{7\pi }{6},n\in z$