# Confusion with very basic algebra

1. Sep 28, 2006

### quasar987

I'm trying to find the points t in (0,2$\pi$) such that sint=sin4t. So I use the fact that sinA=sinB <==> A=B+2n$\pi$ ($n\in\mathbb{Z}$), which yields t=2n$\pi$/3 ($n\in\mathbb{Z}$). The only solutions of this in (0,2$\pi$) are 2$\pi$/3 and 4$\pi$/3.

However, there are 7 intersection points, says the "indirect method" of finding the zeros in (0,2$\pi$) of sint-sin4t = sint+sin(-4t) =2sin(3t/2)cos(5t/2).

The number 7 is also confirmed by my calculator. So why doesn't the direct method using only the properties of sine, give the correct answer?!?!

Last edited: Sep 28, 2006
2. Sep 28, 2006

### StatusX

sinA=sinB <=> A=B+2npi or A=(2n+1)pi-B. Remember that sin(x) is periodic with period 2pi, but not one-to-one on a given period (ie, there are points within a single period with the same sin).

3. Sep 28, 2006

### quasar987

Hey, you're right! Thanks StatusX.