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
 A: 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 $x=0$?
A: 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.
A: Another way to look:
\begin{eqnarray*}
x^{4}-2 x^{2} \cos^{2}\left(\frac{\pi x}{2}\right) +1 &=& x^{4}-2x^2+1 + 2 x^{2} \cos^{2}\left(\frac{\pi x}{2}\right) \\
&=& \left(\frac{x^{2}-1}{x \sqrt{2}}\right)^{2} + \cos^{2}\left(\frac{\pi x}{2}\right)
\end{eqnarray*}
Now $ \left(\frac{x^{2}-1}{x \sqrt{2}}\right)^{2} + \cos^{2}\left(\frac{\pi x}{2}\right)=0$ implies,
\begin{equation}
0 \ge - \left(\frac{x^{2}-1}{x \sqrt{2}}\right)^{2} = \cos^{2}\left(\frac{\pi x}{2}\right) \ge 0
\end{equation}
Bounding from both side means, it has to be equality. 
\begin{equation}
0 = - \left(\frac{x^{2}-1}{x \sqrt{2}}\right)^{2} = \cos^{2}\left(\frac{\pi x}{2}\right)  = 0
\end{equation}
This is satisfied only with $x=\pm 1$.
A: If $f(x)\ge0$ and $g(x)\ge0$, for all $x$, then
$$
f(x)+g(x)=0
\qquad\text{if and only if}\qquad
f(x)=0\text{ and }g(x)=0
$$
Take $f(x)=(x^2-1)^2$. This equals zero only at $-1$ and $1$.
If $g(x)=2x^2\cos^2\bigl(\frac{\pi x}{2}\bigr)$, is $g(1)=0$ or $g(-1)=0$? No other values of $x$ can make $f(x)+g(x)=0$.
A different strategy is to note that $0\le\sin^2(\pi x/2)\le1$, so that $-2x^2\le-2x^2\sin^2(\pi x/2)$ and therefore
$$
x^4-2x^2+1\le x^4-2x^2\sin^2(\frac{\pi x}{2})+1
$$
If the right-hand side is $0$, then also $x^4-2x^2+1=(x^2-1)^2$ must be $0$, which implies $x=1$ or $x=-1$. Then it's just a matter of checking whether these values are solutions of the given equation.
A: Solving for $x^2$ we have
$$
x^2 = \frac 12\left(2\sin^2\left(\frac{x \pi}{2}\right)\pm\sqrt{4\sin^4\left(\frac{x \pi}{2}\right)-4}\right)
$$
now we know that $-1\le \sin \left(\frac{x \pi}{2}\right)\le 1$ so the only real possible solution is for $x = \pm 1$
A: Notice that if $x_0$ is a root, then so is $-x_0$. Can you see what the sum is now?
