Question

Find all solutions to the equation in the interval [0, 2π).

sin 2x - sin 4x = 0

Answer 1

Are those exponents on the sine functions, or coefficients on the x's?

Answer 2

\[\large \sin 2x- \sin 4x=0\]Like that? :D

Answer 3

Yes!

Answer 4

First of all, we'll need to apply the Sine Double Angle Formula to the second term.

Recall this identity:\[\large \sin 2 \theta=2 \sin \theta \cos \theta\]

In our problem, we have \(2\theta=4x\), understand how we're going to apply this rule? :D

Answer 5

\[\large \sin 4x \quad = \quad \sin 2(2x)\]Applying the rule gives us,\[\large 2 \sin(2x) \cos (2x)\]

So our problem now becomes,\[\large \sin(2x)-\sin(4x)=0 \\\large \sin(2x)-2\sin(2x)\cos(2x)=0\]

Answer 6

Bah, sorry if you were typing it out :D haha

Answer 7

From here, you might notice that each term has a sin(2x) in it. We can factor that out.

Answer 8

\[\large \sin(2x)\left[1-2\cos(2x)\right]=0\]

Answer 9

Im so confused...Ive been having a lot of trouble with this lesson

Answer 10

\[\sin2x-\sin4x=\sin2x-\sin(2x+2x)=\sin2x-[\sin2xcos2x+\cos2xsin2x]=\]
\[\sin2x-2\sin2xcos2x=\sin2x(1-2\cos2x)=0\]
\[\sin2x=0 or 1-2\cos2x=0\]
If sin2x=0, 2x= 0, 180, 360, 540 so x = 0, 90, 180, 270
1-2cos2x=0
cos 2x=1/2
2x= 60, 300, 420, 660
x=30,150, 210, 330

Answer 11

Mert applied the Angle Sum Identity for Sine, maybe apply that instead of the Sine Double Angle if it makes more sense to you. :D

Answer 12

Thank you so much, the Angle Sum Identity makes more sense to me than the Sine Double Angle does