The problem was:
It can be shown that the below equation has a unique root in the interval $(0,2)$.
Can you find it?
$$ \sqrt{2 + \sqrt{2 - \sqrt{2 + x}}} = x $$
Solution
This can be solved neatly using trigonometry!
Set $x = 2\cos \theta$ for some $\theta \in \left[0, \dfrac{\pi}{2}\right]$
Now $$\sqrt{2+x} = \sqrt{2 + 2\cos \theta} = 2 \cos \frac{\theta}{2}$$ using the double angle formula $1 + \cos 2y = 2 \cos^2 y$.
Then use $1 - \cos 2y = 2 \sin^2 y$ and $ \cos y = \sin (\pi/2 - y)$, and you can find out what $\theta$ is.
It can be shown that the below equation has a unique root in the interval $(0,2)$.
Can you find it?
$$ \sqrt{2 + \sqrt{2 - \sqrt{2 + x}}} = x $$
Solution
This can be solved neatly using trigonometry!
Set $x = 2\cos \theta$ for some $\theta \in \left[0, \dfrac{\pi}{2}\right]$
Now $$\sqrt{2+x} = \sqrt{2 + 2\cos \theta} = 2 \cos \frac{\theta}{2}$$ using the double angle formula $1 + \cos 2y = 2 \cos^2 y$.
Then use $1 - \cos 2y = 2 \sin^2 y$ and $ \cos y = \sin (\pi/2 - y)$, and you can find out what $\theta$ is.
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