Math - Trig Identity?

Question

sin 2x + sin 4x = 0

Can someone please help me solve this?
And show the steps too?

Answer 1

Assuming a restriction in the interval from [0, 2pi),

sin2x + sin4x = 0

Recall the identity, sin2y = 2sinycosy. All we have to do is apply it to sin4x. It isn't immediately obvious, but I'll make it obvious. Since 4x = 2(2x), we have

sin4x = sin[2(2x)]

And applying the double angle identity,
sin[2(2x)] = 2sin(2x)cos(2x).

Back to our equation:

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

Factoring out sin(2x), we get

sin(2x) [1 + 2cos2x] = 0

Now we equate each of these two 0; that is, sin(2x) = 0,
1 + 2cos2x = 0. Let's solve these individually and merge our answers together.

sin(2x) = 0

Sine is equal to 0 at two points on the unit circle: 0 and pi. Since we have 2x, we have to effectively double our values, though.

2x = {0, pi, 2pi, 3pi}

Dividing each solution by 2, we get

x = {0, pi/2, pi, 3pi/2}

For our second equation,
1 + 2cos2x = 0
2cos2x = -1
cos(2x) = -1/2

Cosine is equal to -1/2 at the two points 7pi/6 and 11pi/6. But again, we have to double the values by adding 2pi to each of these two solutions.

2x = {7pi/6, 11pi/6, 7pi/6 + 2pi, 11pi/6 + 2pi}
2x = {7pi/6, 11pi/6, 19pi/6, 23pi/6}

Dividing both sides by two (or multiplying by one half), we get

x = {7pi/12, 11pi/12, 19pi/12, 23pi/12}

So our COMBINED solutions are:

x = {0, pi/2, pi, 3pi/2, 7pi/12, 11pi/12, 19pi/12, 23pi/12}

Answer 2

Solve for x for

sin 2x + sin 4x = 0

sin(2x) + 2sin(2x)cos(2x) = 0
sin(2x)[1 + 2cos(2x)] = 0

sin 2x = 0
2x = 0 + πn, n an integer
x = 0 + πn/2

cos 2x = -1/2
2x = 2π/3 + 2πn, 4π/3 + 2πn
x = π/3 + πn, 2π/3 + πn

The answer is:

x = {0, π/3, π/2, 2π/3} + πn where n is an integer