What are the solutions to the equation sin x + sin 2x = 0 on [0, 2p)?

Question

I'm trying to study for a huge online trig test and the videos and sample problems they give you don't help at all, so can you please show the steps.

Answer 1

sinx + sin2x = 0
sinx + 2sinxcosx = 0
sinx(1+2cosx) = 0

The equation is 0 when sinx is 0 (ie x=0, x=pi)

and the equation is 0 when 1+2cosx=0 or cosx=-1/2 (ie x= 2pi/3, x=4pi/3)

So the equation holds at x=0, 2pi/3, pi, 4pi/3

Answer 2

Recall:
sin 2x = 2 sin x cos x.
So
sin x + 2 sin x cos x = 0.
sin x(1 + 2 cos x ) = 0.
So sin x = 0, and x = 0 or π
or
cos x = -1/2
x = 2π/3 or 4π/3.

Answer 3

use trig identities to rewrite as
sin x + 2 sin x cos x = 0
sin x (1 + 2 cos x) = 0
sin x = 0 and 1 + 2 cos x = 0, or cos x = -1/2
using unit circle,
sin x = 0 at x = 0, pi
cos x = -1/2 at x = 2pi/3, 4pi/3

Answer 4

sin x + sin 2x = 0
sin x + 2sin x cos x = 0
sin x ( 1 + 2cos x ) = 0
sin x = 0 or ( 1 + 2cos x ) = 0
sin x = 0 or cos x = -1/2
x = 180 , 0 or 120 , 240

Answer 5

sin x + sin 2x = 0
sin 2x = 2sinxcosx (trig identity)
sinx + 2sinxcosx = 0
sinx (1 + 2cosx) = 0
sinx = 0 or cosx = -0.5
sinx = 0 when x = 0 or pi
cosx = -.5 when x = (2pi)/3 or (4pi)/3
so answer:
x = (0, 2pi/3, pi, 4pi3)

Answer 6

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

double angle formulas:
sin(2x) = sinx cosx + cosx sin x = 2 sinx cosx

sinx + 2 sinx cosx = 0

factor out sin(x)
sinx (1 + 2cosx) = 0

sinx = 0
x = 0 and pi

1 + 2cosx = 0
2cosx = -1
cosx = -1/2
x = 2pi/3 and 4pi/3

Answer 7

sinx+sin2x=0
sinx +2 sinx cosx=0
sinx=0, x=0,2pi
ORcosx=-1/2=-cos(pi/3),x=2pi/3,7pi/3. answer

Answer 8

3sin2x+cos2x=0 ie, 3.sqrt(a million-cos^2(2x))+cos2x=0 ie, 3.sqrt(a million-cos^2(2x))=-cos2x ie, 9(a million-cos^2(2x))=cos^2(2x) ie, 9-9cos^2(2x)=cos^2(2x) ie, 10cos^2(2x)=9 ie, cos^2(2x)=0.9 ie, cos(2x)=0.ninety 5 ie, 2x=18.40 3 levels ie, x=9.22degrees