# Finding all solutions of to the equation

• Apr 3rd 2009, 01:45 PM
yeunju
Finding all solutions of to the equation
Find all solutions to the equation in the interval [0, 2pi)

Problem: sin 2x = -sin x

My attempt:
I'm thinking of making it equal to 0

sin2x - sin x = 0
factor out sin

sin(2x-1) = 0
2x-1= 0
x =1/2

I dont know if i'm going the right direction .

sin(1/2) ?
on unit circle sin 1/2 = pi/6, 5pi/6, 11pi/6
• Apr 3rd 2009, 02:21 PM
skeeter
Quote:

Originally Posted by yeunju
Find all solutions to the equation in the interval [0, 2pi)

Problem: sin 2x = -sin x

My attempt:
I'm thinking of making it equal to 0

sin2x - sin x = 0
factor out sin

sin(2x-1) = 0
2x-1= 0
x =1/2

I dont know if i'm going the right direction .

sin(1/2) ?
on unit circle sin 1/2 = pi/6, 5pi/6, 11pi/6

sorry to tell you this, but you are way off ...

$\sin(2x) = -\sin{x}$

$\sin(2x) + \sin{x} = 0$

use the double angle identity for sine ...

$2\sin{x}\cos{x} + \sin{x} = 0$

factor ...

$\sin{x}(2\cos{x} + 1) = 0$

set each factor equal to 0 ...

$\sin{x} = 0$ ... $x = 0 \, , x = \pi
$

$\cos{x} = -\frac{1}{2}$ ... $x = \frac{2\pi}{3} \, x = \frac{4\pi}{3}$