<s>[INST] 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
 [/INST] Hint:  This hint also handles complex roots.  Let $a,b,c\in\mathbb{C}$.  Prove that, for a complex number $z$, $x:=z$ is a solution to $$x^4-a\,x^2\,\sin^2(bx)+c=0$$ if and only if $x:=-z$ is a solution of the above equation.
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