Question

How do you solve `sin(x) + sin 2(x) = 0`?

Answer 1

`x={2n\pi}/3` or `x=(2n+1)\pi`.

Explanation:

Use the trigonometric sum product relation:

`\sin(u) +\sin(v) = 2 \sin({u+v}/2)\cos({u-v}/2)`

`\sin(2x) +\sin(x) = 0 = 2 \sin({3x}/2)\cos({x}/2)`

Then either factor may he zero.

#\sin({3x}/2)=0; {3x}/2n\pi; x={2n\pi}/3#

Or:

#cos(x/2)=0; x/2={(2n+1)\pi}/2; x=(2n+1)\pi#

In both cases `n` is any integer.

Answer 2

`x = kpi`
`x = +- (2pi)/3 + 2kpi`

Explanation:

Replace in the equation (sin 2x) by 2sinx.cos x -->
sin x + 2sin x.cos x = 0
sin x(1 + 2cos x) = 0
Either factor should be zero.

a. sin x = 0
x = 0, `x = pi`, `x = 2pi` -->
General: `x = kpi`
b. 2cos x + 1 = 0 --> `cos x = -1/2`
Trig table and unit circle give 2 solutions:
`x = +- (2pi)/3`
General answers:
`x = +- (2pi)/3 + 2kpi`