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

2 Answers
Mar 26, 2018

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.

Mar 27, 2018

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