Question

Solve the following equations.
$8cosxcos2xcos4x=\frac{sin6x}{sinx}.$

Answer 1

$8cosxcos2xcos4x=\frac{sin6x}{sinx}$

$\Rightarrow 8sinx.cosx.cos2x.cos4x=sin6x$

$\Rightarrow 2sinx.cosx.4cos2x.cos4x=sin6x$

$\Rightarrow 4sin2xcos2xcos4x=sin6x$

$\Rightarrow 2(2sin2x.cos2x)cos4x=sin6x$

$\Rightarrow 2sin4x.cos4x=sin6x$

$\Rightarrow sin8x-sin6x=0$

$\Rightarrow 2cos7x.sinx=0$

$\therefore cos7x=0sinx=0$

$7x=(2x+1)\pi /2x=m\pi$

$x=(2x+1)\pi /4mϵz$

$xϵz$