Question

What are all the solutions to the equation sin 2x = 2sin x in the interval [0, 2π)?

{π}
{0, π}
{pi over 2}
{pi over 2, 3 times pi over 2}

Answer 1

`sin(2x) = 2sin(x) \\ 2sin(x)cos(x) = 2sin(x) \\ sin(x)cos(x) = sin(x) \\ sin(x)cos(x) - sin(x) = 0 \\ \sin(x) (cosx - 1) = 0 \\ \sin(x) = 0 \\ x = \pi \: or \: x = 0 \\ cos(x) = 1 \\ x = 0 \: or \: x = \pi \\ \\ s`

`s = > {0 < = > \pi}`