Q:

How do I solve $(\cos2x+1)^2=1/2$?

$x$ belongs to $$[0,\pi/2[$$
$$(\cos(2x+1))^2 = \frac{1}{2}$$
I tried to find $x$ using
$$2x+1=\frac{\pi}{4} +\frac{n\pi}{2}
\qquad\text{or}\qquad
2x+1=\frac{3\pi}{4}  + n\pi$$
but I didn't find my answer in MCQ which is
$$\frac{3\pi-4}{8}
\qquad\text{and}\qquad
\frac{5\pi-4}{8}.$$
I just need a hint on how to solve such equation.

A:

Your inital step is right, indeed we have that
$$(\cos(2x+1))^2 =\frac12 \iff \cos(2x+1)=\pm\frac{\sqrt 2}2 \iff 2x+1=\frac \pi 4+k\frac \pi 2$$
then
$$x=\frac \pi 8+k\frac \pi 4-\frac12$$
and since $x\in [0,\pi/2[$ we obtain
$$x_1=\frac 3 8\pi-\frac12, x_2=\frac 5 8\pi-\frac12$$

