<s>[INST] 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.
 [/INST] $$\implies\cos2(2x+1)=2\cos^2(2x+1)-1=0$$
$2(2x+1)= \dfrac{(2n+1)\pi}2 $ where $n$ is any integer
Now $1\le2x+1\le1+\pi$
If $1\le\dfrac{(2n+1)\pi}4\le1+\pi\iff\dfrac4\pi-2\le n\le\dfrac4\pi\implies n=0,1$
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