General Solutions

• Jun 4th 2009, 09:00 PM
xwrathbringerx
General 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 PM
Amer
do you know the formula

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

so

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

so sin2x=0 or cosx=0

sin2x=0

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

$cosx=0.....x=\frac{n\pi}{2}$
• Jun 4th 2009, 09:59 PM
mr fantastic
Quote:

Originally Posted by xwrathbringerx
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?

A simple approach:

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

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

Case 2: $3x = (\pi - (-x)) + 2 n \pi = x + (2n + 1) \pi$. The solutions here are a subset of those in Case 1.
• Jun 5th 2009, 06:46 AM
xwrathbringerx
hmmmm can x = (pi*n)/(3+(-1)^n)?
• Jun 5th 2009, 07:15 AM
Amer
yeah it can be I think (Thinking)
• Jun 5th 2009, 04:24 PM
mr fantastic
Quote:

Originally Posted by xwrathbringerx
hmmmm can x = (pi*n)/(3+(-1)^n)?

This is a complicated way of saying $x = \frac{n \pi}{2}$ where n is an integer.