Vtyjkyuk

My try:
$$x^4-2x^2sin^2(fracpi x2)+1=0\x^4+1=2x^2left (1-cos^2left(fracpi x2right)right )\(x^2-1)^2=-2x^2cos^2left(fracpi x2right)\(x^2-1)^2+2x^2cos^2left(fracpi x2right)=0\x^2-1=0,textand, 2x^2cos^2left(fracpi x2right)=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 $$x^2=0implies x=0$$ or $$cosfracpi x2=0implies x=frac2picdotleft(fracpi2+pi kright)$$ for some integer $k$.

But can $x=0$?

Hint: This hint also handles complex roots. Let $a,b,cinmathbbC$. 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.

Another way to look:

begineqnarray*
x^4-2 x^2 cos^2left(fracpi x2right) +1 &=& x^4-2x^2+1 + 2 x^2 cos^2left(fracpi x2right) \
&=& left(fracx^2-1x sqrt2right)^2 + cos^2left(fracpi x2right)
endeqnarray*
Now $ left(fracx^2-1x sqrt2right)^2 + cos^2left(fracpi x2right)=0$ implies,
beginequation
0 ge - left(fracx^2-1x sqrt2right)^2 = cos^2left(fracpi x2right) ge 0
endequation

Bounding from both side means, it has to be equality.
beginequation
0 = - left(fracx^2-1x sqrt2right)^2 = cos^2left(fracpi x2right) = 0
endequation

This is satisfied only with $x=pm 1$.

If $f(x)ge0$ and $g(x)ge0$, for all $x$, then
$$
f(x)+g(x)=0
qquadtextif and only ifqquad
f(x)=0text and g(x)=0
$$
Take $f(x)=(x^2-1)^2$. This equals zero only at $-1$ and $1$.

If $g(x)=2x^2cos^2bigl(fracpi x2bigr)$, 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 $0lesin^2(pi x/2)le1$, so that $-2x^2le-2x^2sin^2(pi x/2)$ and therefore
$$
x^4-2x^2+1le x^4-2x^2sin^2(fracpi x2)+1
$$
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.

Solving for $x^2$ we have

$$
x^2 = frac 12left(2sin^2left(fracx pi2right)pmsqrt4sin^4left(fracx pi2right)-4right)
$$

now we know that $-1le sin left(fracx pi2right)le 1$ so the only real possible solution is for $x = pm 1$

Notice that if $x_0$ is a root, then so is $-x_0$. Can you see what the sum is now?