Answer
$x=0, \dfrac{\pi}{2},\dfrac{3 \pi}{2}$
Work Step by Step
We have $\sin( x +\pi/2) =\sin x \cos (\pi/2) +\cos x \sin (\pi/2) =0+\cos x+\cos x$
After simplifying, we get $\cos x-(\cos x)^2 =0$
or, $ \cos x (1-\cos x )=0$
When $\cos x=0 \implies x=\dfrac{\pi}{2},\dfrac{3 \pi}{2}$
and when $1- \cos x -1=0 \implies x=0$
Thus, we have: $x=0, \dfrac{\pi}{2},\dfrac{3 \pi}{2}$
