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,

$\sin^2x=\dfrac{1-\cos2x}2$

Then

$\sin^2x=\dfrac{1-\cos2x}2=\dfrac1{36}\implies\cos2x=\dfrac{17}{18}$

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

$2x=\cos^{-1}\dfrac{17}{18}+2n\pi$

$\implies x=\dfrac12\cos^{-1}\dfrac{17}{18}+n\pi$

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

$x=\dfrac12\cos^{-1}\dfrac{17}{18}$

or

$x=\dfrac12\cos^{-1}\dfrac{17}{18}+\pi$