Question

solve sin^2 x = sin x for [0,2pi]

Answer 1

graph them both and see where the overlap in the interval of [0, 2pi]

Answer 2

trying to find a more accurate way currently gimme a second

Answer 3

\(\bf sin^2(x) = sin(x)\implies sin^2(x)-sin(x) = 0\implies sin(x)[sin(x)-1]=0\)

solve for "x" using the inverse sine function