<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] 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.
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