Question

Solve the following equations.
$si{n}^{2}x+si{n}^{2}2x-si{n}^{2}3x-si{n}^{2}4x=0.$

Answer 1

$si{n}^{2}x+si{n}^{2}2x-si{n}^{2}3x-si{n}^{2}4x=0$

Using,$si{n}^{2}A-si{n}^{2}B=sin(A+B),sin(A-B)$ we can resorite the above

$\Rightarrow si{n}^{2}x-si{n}^{2}3x+si{n}^{2}2x-si{n}^{2}4x=0$

$\Rightarrow sin4xsin(-2x)+sin6x.sin(-2x)=0$

$\Rightarrow sin(-2x)[sin6x+sin4x]=0$

$\Rightarrow sin(-2x)(2sin5x.cos2)=0$

$\therefore sin(-2x)=0$ or $sin5x=0$ or $cosx=0$

$x=x\pi /2$ or $5x=m\pi$ or $x=(2k+1)\pi /2$

$x=m\pi /5$

$mϵ$ integers $mϵ$ integer $kϵ$ integers.