Fun with trigo

By using AM-GM Inequality, we will get the RHS to be $\frac{1}{2}(y+\frac{1}{y})≥\sqrt{y×\frac{1}{y}}\Rightarrow \sqrt{y+\frac{1}{y}} ≥\sqrt{2}$ While the LHS may be written into $\sqrt{1+\sin 2x}$ , or simply $-\sqrt{2} ≤ \sin(x)+\cos(x) ≤\sqrt{2}$ The only point when the two equals is $\sqrt{2}$ , which, in the range given, only happens when $x=\frac{\pi}{4}$

The solution $\frac{5 \pi}{4}$ is invalid, because it doesn't lie in the interval $[0,\pi]$ .

The solution $\frac{3 \pi}{4}$ is invalid, because it would mean the left hand side becomes zero.

This results in the equation $\sqrt{y + \frac{1}{y}} =0$ , which has no real solutions (the sum of a number and its multiplicative inverse can never become zero in $\mathbb{R}$ , you are welcome to check this algebraically.).

The solution $\frac{7 \pi}{8}$ is invalid, because the absolute value of the cosine is clearly greater than that of the sine. However, since cosine is negative for this value and sine positive, this would result in a negative left hand side, which results in no real solutions for the root.

Hence $\frac{ \pi}{4}$ is the only possible solution listed (but not the only possible solution!).