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

Question

12673 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-11T08:54:01

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

$2sin x*cos x+sin x=0$ 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 $0, 2pi/3, pi, 4pi/3$ .

How do you solve 2sin x cos x +sin x=02sin x cos x +sin x=0 for x in the interval [0,2pi)?