General Solutions

xwrathbringerx · #1 $Old$ June 4th, 2009, 10:00 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 · #2 $Old$ June 4th, 2009, 10:58 PM

do you know the formula

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

so

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

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 · #3 $Old$ June 4th, 2009, 10:59 PM

Quote:

Originally Posted by xwrathbringerx $View Post$

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 · #4 $Old$ June 5th, 2009, 07:46 AM

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

Amer · #5 $Old$ June 5th, 2009, 08:15 AM

yeah it can be I think

mr fantastic · #6 $Old$ June 5th, 2009, 05:24 PM

Quote:

Originally Posted by xwrathbringerx $View Post$

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.