<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] Assuming $x\in \Bbb R.$ For brevity let $c=\cos x$ and $s=\sin x.$ Let $p=c^2s^2.$ We have $$ c^8+s^8=\frac {3}{8}\iff$$ $$ 1=(c^2+s^2)^4=(c^8+x^8)+c^2s^2(4c^4+6c^2s^2+4s^4)=$$ $$=\frac {3}{8}+c^2s^2(4(c^2+s^2)^2-2c^2s^2)=$$ $$=\frac {3}{8}+c^2s^2(4-2c^2s^2)\iff$$  $$\iff(0\leq p\leq 1\land \frac {5}{16}=2p-p^2)$$  $$\iff p=1- \sqrt {11}\;/4\iff$$ $$\iff |\sin 2x|=\sqrt {4p}=\sqrt {4-\sqrt {11}}.$$
Note: The 2nd, 3rd, and 4th displayed lines are a single sentence that is true iff $c^8+x^8=\frac {3}{8}.$
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