General Solutions

**xwrathbringerx** · June 4th 2009, 09:00 PM

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?

**Amer** · June 4th 2009, 09:58 PM

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

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

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

**mr fantastic** · June 4th 2009, 09:59 PM

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.

**xwrathbringerx** · June 5th 2009, 06:46 AM

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

**Amer** · June 5th 2009, 07:15 AM

yeah it can be I think

**mr fantastic** · June 5th 2009, 04:24 PM

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.

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