# Trigonometric equations - strange results

1. Apr 24, 2004

### repugno

Greetings all,

I am getting strange results when solving this trig equation. I seem to be able to calculate 4 out 7 of the correct angles but how do i calculate the others? Maybe my method is wrong...

A = theta

2sin2A = tanA,

considering identities sin2A = 2sinAcosA and tanA = sinA/cosA

2(2sinAcosA) = sinA/cosA

4sinAcos^2A = sinA

dividing by sinA both sides

4cos^2A = 1
cos^2A = 1/4
cosA = 1/2
cosA = -1/2

A = 60
A = 300
A = 120
A = 240

missing angles 0, 180 and 360 ??

Any help would be much appreciated, thanks

2. Apr 24, 2004

### arildno

When dividing with sin(A), you have assumed sin(A) not equal to zero.

3. Apr 25, 2004

### recon

I think that you have to illustrate that

$$4sin(A)cos^2(A) = sin(A)$$

$$4sin(A)cos^2(A) - sin(A) = 0$$

So

$$sin(A)(4cos^2(A) - 1) = 0$$

4. Apr 25, 2004

### Hurkyl

Staff Emeritus
Or, alternatively, you have to break the problem up into two cases; case 1 is where sin A is 0, and case 2 is where sin A is not zero (and thus you can divide by sin A)

But whatever you do, the point we're making is that, in general, you cannot divide by something that may be zero in your problem.