Question

(1 point) Solve the equation in the interval [0,2π]. If there is more than one solution write them separated by commas. (sin(x))2=1/36

Answer 1

Recall the double angle identity,

`\sin^2x=\frac{1-\cos2x}2`

Then

`\sin^2x=\frac{1-\cos2x}2=\frac1{36}\implies\cos2x=(17)/(18)`

`\cos x` has a period of
`2\pi`, so that
`\cos x=\cos(x+2n\pi)` for any integer
`n`. This means

`2x=\cos^(-1)(17)/(18)+2n\pi`

`\implies x=\frac12\cos^(-1)(17)/(18)+n\pi`

We get solutions of this form in the interval
`[0,2\pi]` for
`n=0` and
`n=1`, giving

`x=\frac12\cos^(-1)(17)/(18)`

or

`x=\frac12\cos^(-1)(17)/(18)+\pi`