Sum of roots of equation $x^4 - 2x^2 \sin^2(\displaystyle {\pi x}/2) +1 =0$ is My try:
$$x^4-2x^2\sin^2(\frac{\pi x}{2})+1=0\\x^4+1=2x^2\left (1-\cos^2\left(\frac{\pi x}{2}\right)\right )\\(x^2-1)^2=-2x^2\cos^2\left(\frac{\pi x}{2}\right)\\(x^2-1)^2+2x^2\cos^2\left(\frac{\pi x}{2}\right)=0\\x^2-1=0\,\text{and}\, 2x^2\cos^2\left(\frac{\pi x}{2}\right)=0$$
I am stuck , I am confused now what to do now

HINT:
I'm sure you can solve the equation $x^2-1=0$.
For the second equation, you have that either $$x^2=0\implies x=0$$ or $$\cos\frac{\pi x}2=0\implies x=\frac2\pi\cdot\left(\frac\pi2+\pi k\right)$$ for some integer $k$.
But can $x=0$?
