# Homework Help: Simple Trig Question

1. Feb 15, 2012

### kscplay

1. The problem statement, all variables and given/known data
-sin x = √3 * cos x
where x is [0,2π]

2. Relevant equations

3. The attempt at a solution
Would it be wrong to square both sides and then factor?
sin2 x = 3cos2 x
1-cos2 x = 3cos2 x
0= 4cos2 x - 1
0 = (2cos x -1)(2cos x + 1)
Now I solve the factors (2 solutions for each)

Since I've squared both sides, haven't I introduced extra solutions that are not part of the original problem? Will all 4 resulting solutions be correct? Thanks.

2. Feb 15, 2012

### Shootertrex

Well, when you square root something, you will get a positive and a negative answer, which you already know. Since you defined x as [0,2π], there are no negative numbers in that interval. Therefore, you may disregard the negative answers for this interval, which should leave you with two answers.

3. Feb 15, 2012

### tiny-tim

hi kscplay!
yes, since you will also be finding the solutions to sin x = √3 * cos x

the same x can't be a solution to both (unless sinx = 0, of course)

i think the only way to decide which solutions to reject is to actually check each one

4. Feb 15, 2012

### HallsofIvy

But the simplest way to solve that equation is to divide both sides by -cos(x):
$$\frac{sin(x)}{cos(x)}= tan(x)= -\sqrt{3}$$