If $\sin^8(x)+\cos^8(x)=48/128$, then find the value of $x$? If $$\sin^8(x)+\cos^8(x)=48/128,$$ then find the value of $x$? I tried this by De Moivre's theorem:

$$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(\theta)=e^{in\theta}$$

But could not proceed further please help. 

\begin{align}
\sin^8 x+\cos^8 x&=(\sin^4 x+\cos^4)^2-2\sin^4 x\cos^4 x\\
&=\left[(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x\right]^2-\cfrac {\sin^4 2x} 8\\
&=(1-\cfrac {\sin^2 2x}2)^2--\cfrac {\sin^4 2x} 8
\end{align}
Let $t=\sin^2 2x$, we have
$$(1-\cfrac {t}2)^2-\cfrac {t^2}8=\cfrac {48}{128}\Leftrightarrow t^2-8t+5=0,$$ which implies
$$\sin^2 2x=4-\sqrt{11}  \hspace{1in}\text{(with the other invalid root discarded)}$$
hence $$ x=\pm \cfrac {\sin^{-1}(\sqrt{4-\sqrt{11}})}2$$or more precisely,
$$x=\pm\cfrac {\sin^{-1}(\sqrt{4-\sqrt{11}})}2+k\pi,\>k\in \mathbb{Z}$$
