Question

How do you solve `sin x + 2sin x cos x = 0`?

Answer 1

`x in {color(red)(npi), color(blue)((2pi)/3+n(2pi)), color(green)((4pi)/3+n(2pi))} AA n in ZZ`

Explanation:

`sin(x)+2sin(x)cos(x) = 0`

Factoring the left side:
`(sin(x))*(1+2cos(x))=0`

Either `color(red)(sin(x) = 0)` or `color(orange)(1+2cos(x)=0)`

If `color(red)(sin(x)=0)`
`rarr x in {color(red)(npi)} AA n in ZZ`
(i.e. `x` is a multiple of `pi` radians)

If `color(orange)(1+2cos(x)=0)`
`rarr cos(x) = -1/2`
withing the interval `[0,2pi]`
`x= color(blue)((2pi)/3)` or `color(green)(x=(4pi)/3)`
or, more generally:
`x= color(blue)((2pi)/3+n*2pi)` or `color(green)(x=(4pi)/3+n*2pi) AAn in ZZ`