<s>[INST] The value of x satisfying the equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ The value of x satisfying the equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$
(A) $2 \cos 10$
(B) $2 \cos 20$
(C) $2 \cos 40$
(D) $2 \cos 80$
I was dumb enough to square the expression to reach $x^8-4x^6+4x^4+2-x=0$ which is clearly a dead end ;-;
 [/INST] Using the trigonometric identity, $$\cos2x=2\cos^2x-1=1-2\sin^2x$$
$$2\cos2x+2=4\cos^2x$$
We are looking for an angle that allows for the relationship $\sin x=\cos(\frac{\pi}{2}-x)$ in the final step. This is because, working from the inside out, the first gives a cosine, the second gives a sine, so in order for the last one to be a cosine, we must change the sine into a cosine through the above relation.
A simple check/use of intuition finds that C is the answer.
$$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
If $x=2\cos40$,
$$\sqrt{2+\sqrt{2-\sqrt{2+2\cos40}}}=\sqrt{2+\sqrt{2-\sqrt{4\cos^220}}}$$
$$=\sqrt{2+\sqrt{2-2\cos20}}=\sqrt{2+\sqrt{4\sin^210}}=\sqrt{2+2\sin10}$$
$$=\sqrt{2+2\cos80}=2\cos40=x$$
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