# Trigonometric equation

1. Oct 12, 2012

### Feodalherren

1. The problem statement, all variables and given/known data
All solutions in the interval 0 ≤ x < 2∏

Sin 2x = -2cos x

3. The attempt at a solution

I went about it in two different ways.

The first way:

I simply used the double angle formula to get

2sinx cosx + 2cos x = 0

From this we get that whatever makes cos x = 0 will make the whole equation 0.

There are two answers within the interval; ∏/2 and 3∏/2.

This is the correct solution.

When I was studying for my midterm I went about it in a different way though, and I only got one solution. I was wondering what I was doing wrong as it seems to be all legal algebra to me.

2sinx cosx = -2 cos x

divide both sides by cos x

2sinx = -2

divide by 2

sin x = -1

This only gives me one solution, 3∏/2.

What am I doing wrong?

2. Oct 12, 2012

### Staff: Mentor

This is not the right thing to do. When you do this, you will lose whatever solutions correspond to cos(x) = 0.

Instead of dividing through by a variable, factor things.

2sin(x) cos(x) + 2cos(x) = 0

2cos(x)(sin(x) + 1) = 0

And so on.

3. Oct 12, 2012

### Feodalherren

Ok cool so the first way was correct because when I divide by cos x I'm essentially doing illegal algebra because I technically cannot know if cos x = 0?

I think I got it. Thank you!

4. Oct 12, 2012

### Staff: Mentor

Well, you were partly correct in your first way, but you didn't notice that sin(x) = -1 is also a solution. When you factor the expression, both solutions become obvious.

5. Oct 12, 2012

### Feodalherren

I actually did do that in my notes. I didn't include it since it doesn't give me an extra solution :).
Thanks.