The value of x satisfying the equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$

Question

(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 ;-;

Note that $\cos(2\theta)=2\cos(\theta)^2-1$. Replace $x$ with $2\cos (\theta)$ in the right hand side and using this identity, try to infer what $\theta$ must equal. — Suzet, Jun 28 '18 at 05:13
Hint: $\sqrt{2+2\cos\theta} = 2\cos\frac{\theta}{2}$ and $\sqrt{2-2\cos\theta} = 2\sin\frac{\theta}{2} = 2\cos\frac{\pi-\theta}{2}$ — achille hui, Jun 28 '18 at 05:19
The squaring you did was certainly not dumb. It was a reasonable attempt to solve the problem. Maybe there was an onward trick that you did not see or maybe there is another route to a solution. In this case there is another route. — Ross Millikan, Jun 28 '18 at 05:19
See also here: https://math.stackexchange.com/questions/649968 — Michael Rozenberg, Jun 28 '18 at 05:55

Harry Alli · Accepted Answer · 2018-06-28T05:24:12.157

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 $x=2\cos40$, $$\sqrt{2+\sqrt{2-\sqrt{2+2\cos40}}}=\sqrt{2+\sqrt{2-\sqrt{4\cos^220}}}$$ $$=\sqrt{2+\sqrt{2-2\cos20}}=\sqrt{2+\sqrt{4\sin^210}}=\sqrt{2+2\sin10}$$ $$=\sqrt{2+2\cos80}=2\cos40=x$$

Nice solution, but I'm interested in how you would solve this without the answer choices or inferring that $x$ is in trigonometric form. — dcxt, Jun 28 '18 at 05:31

The value of x satisfying the equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$

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