Solve the equation
$$x+\sqrt{2+\sqrt{2+\sqrt{x}}}=2$$
Solution
We set $x =4\cos^2 4a , \; a \in \left(0, \frac{\pi}{8} \right)$. Hence the LHS is transformed into:
$$x+\sqrt{2+\sqrt{2+\sqrt{x}}}=4\cos^2 4a + \sqrt{2+\sqrt{2+2\cos 4a }} =4\cos^2 4a + \sqrt{2+2\cos 2a} = \\4\cos^2 4a + 2\cos a$$
Hence the equation can be now translated as:
$$\begin{aligned}
4\cos^2 4a +2\cos a =2 & \Leftrightarrow 2 \cos^2 4a = 2 \sin^2 \frac{a}{2} \\
&\Leftrightarrow \cos 4a = \sin \frac{a}{2} \\
&\Leftrightarrow 4a = \frac{\pi}{2} - \frac{a}{2} \\
&\Leftrightarrow a = \frac{\pi}{9}
\end{aligned}$$
Hence $\displaystyle x = 4\cos^2 \frac{4\pi}{9}$.
The exercise can also be found in mathematica.gr
$$x+\sqrt{2+\sqrt{2+\sqrt{x}}}=2$$
Solution
We set $x =4\cos^2 4a , \; a \in \left(0, \frac{\pi}{8} \right)$. Hence the LHS is transformed into:
$$x+\sqrt{2+\sqrt{2+\sqrt{x}}}=4\cos^2 4a + \sqrt{2+\sqrt{2+2\cos 4a }} =4\cos^2 4a + \sqrt{2+2\cos 2a} = \\4\cos^2 4a + 2\cos a$$
Hence the equation can be now translated as:
$$\begin{aligned}
4\cos^2 4a +2\cos a =2 & \Leftrightarrow 2 \cos^2 4a = 2 \sin^2 \frac{a}{2} \\
&\Leftrightarrow \cos 4a = \sin \frac{a}{2} \\
&\Leftrightarrow 4a = \frac{\pi}{2} - \frac{a}{2} \\
&\Leftrightarrow a = \frac{\pi}{9}
\end{aligned}$$
Hence $\displaystyle x = 4\cos^2 \frac{4\pi}{9}$.
The exercise can also be found in mathematica.gr
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