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?
 A: 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:
$$\bbox[5px,border:2px solid #C0A000]{x=2\sin^2\left(\frac 12\arcsin \left(\frac {5-2\sqrt 6}{2}\right)\right)}$$
A: Let $x=u^4,\;2-x=v^4$ with the restriction $u,v≥0$, then we have:
$$\begin{align}&\begin{cases}u-v=1\\u^4+v^4=2\end{cases}
\\\implies &v^4+(v+1)^4=2\end{align}$$
Then, using the key substitution $v+\frac 12=y$ leads to:
$$\begin{align}&\left(y-\frac 12\right)^4+\left(y+\frac 12\right)^4=2\\
\implies &16y^4+24y^2-15=0\end{align}$$
Finally, substitute $y^2=z$, you obtain:
$$16z^2+24z-15=0$$
The solution is completed by reversing the steps.
