Question

The number of solutions of $4{\cos }^2(\frac{\pi }{4}-\frac{x}{2})+\sqrt{4{\sin }^{4}x+{\sin }^{2}2x}=0$ in $x\in [0,\pi ]$ is

Answer 1

$4{\cos }^2(\frac{\pi }{4}-\frac{x}{2})+\sqrt{4{\sin }^{4}x+{\sin }^{2}2x}=0$
Since, ${\cos }^2(\frac{\pi }{4}-\frac{x}{2})\ge 0\text{and}\sqrt{4{\sin }^{4}x+{\sin }^{2}2x}\ge 0$

So, the solution of above equation is exist only when
${\cos }^2(\frac{\pi }{4}-\frac{x}{2})=0\text{and}\sqrt{4{\sin }^{4}x+{\sin }^{2}2x}=0$
$\Rightarrow \frac{1}{2}[\cos (\frac{\pi }{2}-x)+1]=0\text{and}4{\sin }^{4}x+4{\sin }^{2}x{\cos }^{2}x=0$
$\Rightarrow \sin x+1=0\text{and}4{\sin }^{2}x=0\Rightarrow \sin x=-1\text{and}\sin x=0$
Which is not possible.

Hence, there is no solution.