Question

Find exact solution of
$\cos( {2x - \frac{\pi}{2} })^{2} = 1 - \sin(2x - \frac{\pi}{2} )$
for
$0 \leqslant x \leqslant 3\pi$
​

Answer 1

For starters, let $y=2x-\dfrac\pi2$.

Then

$\cos^2y=1-\sin y$

$1-\sin^2y=1-\sin y$

$\sin^2y-\sin y=0$

$\sin y(\sin y-1)=0$

so either

$\sin y=0\implies y=n\pi$

or

$\sin y-1=0\implies\sin y=1\implies y=\dfrac\pi2+2n\pi$

for integers $n$. Solving for $x$ gives us

$2x-\dfrac\pi2=\dfrac\pi2+2n\pi\implies x=\dfrac\pi2+n\pi$

or

$2x-\dfrac\pi2=n\pi\implies x=\dfrac\pi4+\dfrac{n\pi}2$

We get the following solutions for $0\le x\le3\pi$:

$x=\dfrac\pi2,\dfrac{3\pi}2,\dfrac{5\pi}2$

or

$x=\dfrac\pi4,\dfrac{3\pi}4,\dfrac{5\pi}4,\dfrac{7\pi}4,\dfrac{9\pi}4,\dfrac{11\pi}4$