Question

How do you solve `2sin x cos x +sin x=0` for x in the interval [0,2pi)?

Answer 1

Solution for `x` is `0, 2pi/3, pi, 4pi/3`.

Explanation:

`2sin x*cos x+sin x=0` can be written as `sin x*(2cos x+1)=0`

This means either `sin x=0` i.e. `x=0, pi` (`2pi` is not included in the interval `[0, 2pi)`

or `(2cos x+1)=0` i.e. `cos x=-1/2` i.e. `x=2pi/3 or 4pi/3`.

Hence solution set for `x` is `0, 2pi/3, pi, 4pi/3`.