Question

Solve the following equation:
$2{\cos }^{2}x-1=\sin 3x.$

Answer 1

$2(1-{\sin }^{2}x)-1=3\sin x-{4\sin }^{3}x$

$2-{2\sin }^{2}x-1-3\sin x+{4\sin }^{3}x=0$

Let, $\sin x=z.$

${4z}^3-{2z}^2-3z+1=0$

$(z-1)({4z}^2+2z-1)=0$

So, $z=1,\sin x=1,x=2n\pi +\frac{\pi }{2}.$

and ${4z}^2+2z-1=0,z=\frac{-2\pm \sqrt{4+16}}{8}=\frac{-1\pm \sqrt{5}}{4}.$

Let, $a=\frac{-1+\sqrt{5}}{4},b=\frac{-1-\sqrt{5}}{4}.$

So, $x=n\pi \pm {(-1)}^n{\sin }^{-1}\left(a\right)$ and $x=n\pi \pm {(-1)}^n{\sin }^{-1}\left(b\right).$