<s>[INST] 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. 
 [/INST] Let $c=cosx$ and $s=sinx$ :
$s^8+c^8= \frac{48}{2^7}*\frac{2}{2}$
$s^8+c^8= \frac{96}{2^8} $
$(\frac{G}{2^4})^2+(\frac{H}{2^4})^2=\frac{96}{2^8}\implies G^2+H^2=96 $
$\implies s=\pm\frac{\sqrt[4]{G}}{2} \wedge  c=\pm\frac{\sqrt[4]{96-G^2}}{2}$

We gonna use: $s^2+c^2=1\rightarrow c^2=1-s^2$:
$\frac{\sqrt{96-G^2}}{4}=1-\frac{\sqrt{G}}{4}$
$\sqrt{96-G^2}=4-\sqrt{G}\rightarrow$ WolframAlpha finds the solution : $G\approx9.7587$
For $\sqrt{96-g^4}=4-g$ we get real solutions : $g_1\approx0.0406-e \vee g_2\approx\pi-0.0177;$ cause $g_1^2\neq G\implies g_2$ gonna be our result of $g^2=G$
.$(g\approx 3.12389354026129)$ 
So: 
$s=\pm\frac{\sqrt{g}}{2}$
We consider for $sinx$ in the 1st quarter:
$\frac{\sqrt{3}}{2}<\frac{\sqrt{g}}{2}<\frac{\sqrt{\pi}}{2}$
$60^\circ<x<62.4^\circ$;cause of $arcsin(\frac{\sqrt{\pi}}{2})$
Direct:

$x=arcsin(\frac{\sqrt{g}}{2})\approx62.1^\circ$


History and altenative version (with different* solution):
All starts with:
$(c^4+is^4)(c^4-is^4)=\frac{6}{2^4}$
$(\frac{A}{2^2})^2+(\frac{B}{2^2})^2=\frac{6}{2^4}\implies A^2+B^2=6 $
$\implies c^4=\frac{A}{4} \wedge  s^4=\frac{\sqrt{6-A^2}}{4}$
Altenative for:
$c=\pm\sqrt[4]{\frac{A}{4}} \wedge  s=\pm\sqrt[4]{\frac{B}{4}}$ with $s^2+c^2=1$ :
$\sqrt{\frac{B}{4}}=1-\sqrt{\frac{A}{4}}$
$6-A^2=[2-\sqrt{A}]^2$
$6-A^2=4-2\sqrt{A}+A$
$0=A^2+A-2\sqrt{A}-2\rightarrow$ WolframAlpha finds the solution : $A\approx1.7049$
So:
$c=\pm\frac{\sqrt[4]{A}}{\sqrt{2}}$
We consider for $cosx$ in the 1st quarter:

$x=arccos(\frac{\sqrt[4]{A}}{\sqrt{2}})\approx36.1^\circ$

*Also: If I would stand by $s=\pm\frac{\sqrt[4]{A}}{\sqrt{2}}\rightarrow x\approx53.9^\circ$ (1st quarter)

Other possibilities via $E+F=(9)6$
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