Find the general solution of sin (3x) + sin (x) = 0. The textbook says the answer is (pi*n)/2 but I can’t seem to get it. Please help?

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- Jun 4th 2009, 09:00 PMxwrathbringerxGeneral Solutions
Find the general solution of sin (3x) + sin (x) = 0. The textbook says the answer is (pi*n)/2 but I can’t seem to get it. Please help?

- Jun 4th 2009, 09:58 PMAmer
do you know the formula

$\displaystyle sinA+sinB=2{sin(\frac{A+B}{2})cos(\frac{A-B}{2})}$

so

$\displaystyle sin3x+sinx=2sin(\frac{4x}{2})cos(\frac{2x}{2})=2si n(2x)cos(x)=0$

so sin2x=0 or cosx=0

sin2x=0

$\displaystyle sin2x=0.....2x=n\pi....x=\frac{n\pi}{2}$

$\displaystyle cosx=0.....x=\frac{n\pi}{2}$ - Jun 4th 2009, 09:59 PMmr fantastic
A simple approach:

$\displaystyle \sin (3x) + \sin (x) = 0 \Rightarrow \sin (3x) = - \sin (x) = \sin (-x)$.

Case 1: $\displaystyle 3x = -x + 2 n \pi$ where n is an integer.

Case 2: $\displaystyle 3x = (\pi - (-x)) + 2 n \pi = x + (2n + 1) \pi$. The solutions here are a subset of those in Case 1. - Jun 5th 2009, 06:46 AMxwrathbringerx
hmmmm can x = (pi*n)/(3+(-1)^n)?

- Jun 5th 2009, 07:15 AMAmer
yeah it can be I think (Thinking)

- Jun 5th 2009, 04:24 PMmr fantastic