Question

If $\sin x\cdot \sin ({60}^{\circ }-x)\sin ({60}^{\circ }+x)=\frac{1}{4},$ then $x=$

Answer 1

$Sinx.sin(60-x).sin(60+x)$

$=sinx\left[cos\right(60-x-60-x)-cos(60-x+60+x)/2$

$=sinx\left[cos\right(-2x)-cos(120\left)\right]/2$

$=sinx[cos2x+1/2]/2$

$=sinx\left[\right(1-2sin²x)+1/2]/2$

$=sinx[3/2-2sin²x]/2$

$=3sinx/4-2sin³x/2$

$=3sinx-4sin³x/4$

$=1/4\left(sin3x\right)$