Find the real roots of the equation: $\sqrt [4]{x}-\sqrt [4]{2-x}=1$ 
Find the real roots of the following equation: $$\sqrt [4]{x}-\sqrt [4]{2-x}=1$$

This is my textbook contest exercise.
$$\sqrt [4]{x}=\sqrt [4]{2-x}+1$$
$$x=\big(\sqrt [4]{2-x}+1\big)^4$$
I know that
$$(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$
So, we will have some cube powers which introduce still $\sqrt [4]{\cdot}$ roots.
This makes more complicated the equation. How can I find the minimal polynomial using a more simple way?

In this answer, I wanted to construct the trigonometric method that requires significantly less computation.

Since $0<x<2$, making the substitution $x=2\sin^2\theta$, where $0<\theta<\frac {\pi}{2}$, we have:
$$\begin{align}&\sqrt {\sin\theta}-\sqrt {\cos\theta}=\frac {1}{\sqrt [4]{2}}\\
\implies&\sin\theta+\cos\theta=\frac{1}{\sqrt 2}+\sqrt {2\sin (2\theta)}\\
\implies&2\sqrt {\sin (2\theta)}+\sin (2\theta)=\frac 12\end{align}$$
Then the substitution $\sqrt {\sin (2\theta)}=u$, leads to:
$$\begin{align}&2u^2+4u-1=0,~u>0\\
\implies &\sin2\theta=\left(\frac {\sqrt 6-2}{2}\right)^2\\
\implies &\theta =\frac 12\arcsin \left(\frac {5-2\sqrt 6}{2}\right)\end{align}$$
Thus, the final answer is:
$$\bbox[5px,border:2px solid #C0A000]{x=2\sin^2\left(\frac 12\arcsin \left(\frac {5-2\sqrt 6}{2}\right)\right)}$$
