For starters, let [tex]y=2x-\dfrac\pi2[/tex].
Then
[tex]\cos^2y=1-\sin y[/tex]
[tex]1-\sin^2y=1-\sin y[/tex]
[tex]\sin^2y-\sin y=0[/tex]
[tex]\sin y(\sin y-1)=0[/tex]
so either
[tex]\sin y=0\implies y=n\pi[/tex]
or
[tex]\sin y-1=0\implies\sin y=1\implies y=\dfrac\pi2+2n\pi[/tex]
for integers [tex]n[/tex]. Solving for [tex]x[/tex] gives us
[tex]2x-\dfrac\pi2=\dfrac\pi2+2n\pi\implies x=\dfrac\pi2+n\pi[/tex]
or
[tex]2x-\dfrac\pi2=n\pi\implies x=\dfrac\pi4+\dfrac{n\pi}2[/tex]
We get the following solutions for [tex]0\le x\le3\pi[/tex]:
[tex]x=\dfrac\pi2,\dfrac{3\pi}2,\dfrac{5\pi}2[/tex]
or
[tex]x=\dfrac\pi4,\dfrac{3\pi}4,\dfrac{5\pi}4,\dfrac{7\pi}4,\dfrac{9\pi}4,\dfrac{11\pi}4[/tex]
