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

Question

Find exact solution of


`\cos( {2x - (\pi)/(2) })^(2) = 1 - \sin(2x - (\pi)/(2) )`
for

`0 \leqslant x \leqslant 3\pi`
​

Answer 1

For starters, let
`y=2x-\frac\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=\frac\pi2+2n\pi`

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

`2x-\frac\pi2=\frac\pi2+2n\pi\implies x=\frac\pi2+n\pi`

or

`2x-\frac\pi2=n\pi\implies x=\frac\pi4+\frac{n\pi}2`

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

`x=\frac\pi2,\frac{3\pi}2,\frac{5\pi}2`

or

`x=\frac\pi4,\frac{3\pi}4,\frac{5\pi}4,\frac{7\pi}4,\frac{9\pi}4,\frac{11\pi}4`