Question

Solve:

$\sin 3x+\cos 2x=0$

Answer 1

Given : $\sin 3x+\cos 2x=0$

Let us simplify,

$\Rightarrow$$\cos 2x=-\sin 3x$

$\Rightarrow$$\cos 2x=–\cos (\frac{\pi }{2}–3x)[S\mathrm{in}ce,\sin A=\cos (\frac{\pi }{2}–A\left)\right]$

$\Rightarrow$$\cos 2x=\cos (\pi –(\frac{\pi }{2}–3x\left)\right)[\sin ce,-\cos A=\cos (\pi –A\left)\right]$

$\Rightarrow$$\cos 2x=\cos (\frac{\pi }{2}+3x)$

$\Rightarrow$ $2x=2n\pi \pm (\frac{\pi }{2}+3x)$

$\Rightarrow$ $2x=2n\pi +(\frac{\pi }{2}+3x)or2x=2n\pi –(\frac{\pi }{2}+3x)$

$\Rightarrow$$3x-2x=-\frac{\pi }{2}–2n\pi or2x+3x=2n\pi –\frac{\pi }{2}$

$\Rightarrow$ $x=-\frac{\pi }{2}–2n\pi or5x=2n\pi –\frac{\pi }{2}$

$\Rightarrow$ $x=-\frac{\pi }{2}\left(1+4n\right)or5x=\frac{\mathrm{\pi }}{2}\left(4n–1\right)$

$\Rightarrow$ $x=-\frac{\pi }{2}\left(4n+1\right)orx=\frac{\mathrm{\pi }}{10}\left(4n–1\right)$

Hence, The solution of the equation is $x=-\frac{\pi }{2}\left(4n+1\right)orx=\frac{\mathrm{\pi }}{10}\left(4n–1\right)$ where $n\in Z$