Question

Solve the following equation:
${\sin }^{4}x+{\cos }^{4}x=\frac{3-\cos 6x}{4}$

Answer 1

$4{({\sin }^{2}x+{\cos }^{2}x)}^2-{8\sin }^{2}x{\cos }^{2}x=3-({4\cos }^{3}2x+3\cos 2x)$

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

Let $\cos 2x=z.$

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

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

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

Either. $z+1=0,z=-1,\cos 2x=-1,\text{[by placing the value of z=cos 2x]}2x=(2n+1)\pi ,x=(2n+1)\frac{\pi }{2}.$

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

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

So, $\cos 2x=a,2x=2n\pi +{\cos }^{-1}\left(a\right),x=n\pi \pm \frac{1}{2}{\cos }^{-1}\left(a\right).$

and,$\cos 2x=b,2x=2n\pi +{\cos }^{-1}\left(b\right),x=n\pi \pm \frac{1}{2}{\cos }^{-1}\left(b\right).$