Does the equation $2\cos^2 (x/2) \sin^2 (x/2) = x^2+\frac{1}{x^2}$ have real solution? 
Do the equation
  $$2\cos^2 (x/2) \sin^2 (x/2) = x^2+\frac{1}{x^2}$$
  have any real solutions?

Please help. This is an IITJEE question.
Here $x$ is an acute angle.
I cannot even start to attempt this question. I cannot understand.  
 A: the right hand of the equation, you have $$x^2 + \frac1{x^2} = \left(x-\frac 1x\right)^2 + 2 \ge 2 \tag 1$$and equality occurs for $$x = \pm 1.$$
on the left hand side, we have $$ \frac12 \sin^2 x = 2\cos^2 (x/2) \sin^2 (x/2) \le \frac12 \tag 2$$ equality for $$\sin x = \pm 1.$$
but $(1)$ and $(2)$ are inconsistent, therefore $$2\cos^2 (x/2) \sin^2 (x/2) = x^2+\frac{1}{x^2} $$ has no real solutions.
A: Observe $x^2+\frac{1}{x^2} \geq 2$, and simplify left hand side using $\sin(2\theta)=2\sin(\theta)\cos(\theta)$.
A: HINT:
We have
$$2\cos^2(x/2)\sin^2(x/2)=\frac12\sin^2x\le \frac12$$
while 
$$x^2+\frac{1}{x^2}\ge 2$$
A: In the same spirit as baharampuri's answer, the equation write $$\frac{\sin ^2(x)}{2}=x^2+\frac{1}{x^2}$$ The lhs is almays $< \frac 12$ while the rhs is $\geq 2$.
For the last point, consider $$y==x^2+\frac{1}{x^2}$$ $$y'=2 x-\frac{2}{x^3}$$ $$y''=2+\frac{6}{x^4}$$ So, $y'=0$ for $x=\pm 1$ and at this point $y=2$. The second derivative test proves that this is a minimum.
A: Hint:
LHS in blue, RHS in green.

A: Alphy says no real roots and the complex ones are
approximately
$ \pm 0.866835 \pm 0.733199 i$
and
$\pm 1.63597 i$.
