# Homework Help: How to solve this trig equation

1. Aug 22, 2012

### lo2

1. The problem statement, all variables and given/known data

Solve this equation:

$cos^2(2x)=0,36$

For $x \in [-\pi;\pi]$

2. Relevant equations

-

3. The attempt at a solution

$cos^2(2x)=0,36 \Leftrightarrow cos(2x)=\sqrt{0,36} \Leftrightarrow 2x=cos^{-1}(\sqrt{0,36})$

And then I am not sure exactly how to proceed... When should I put in the $2p \pi$ where $x \in Z$, to get all of the possible solutions?

Last edited: Aug 22, 2012
2. Aug 22, 2012

### Staff: Mentor

Not true. cos(2x) can also be negative. In your second equation, you took the square root of the right side, but not the left side.
Also, you should simplify √(.36).

3. Aug 22, 2012

### lo2

I corrected the mistake about not taking the square root on either side. So you mean I should put ± in front of the square root?

4. Aug 22, 2012

### SammyS

Staff Emeritus
Yes, use the ± .

5. Aug 22, 2012

### Staff: Mentor

The domain for x is restricted to [$-\pi, \pi$], so you're going to get only a handful of solutions.

6. Aug 23, 2012

### lo2

Ok I have come up with this solution:

$\frac{cos^{-1}(\pm \sqrt{0,36})}{2}+p\pi$

Where the solutions are: $cos^{-1}(\sqrt{0,36})-\pi, cos^{-1}(-\sqrt{0,36}), cos^{-1}(\sqrt{0,36}), cos^{-1}(-\sqrt{0,36})+\pi$

Since the solutions have to be in the interval of -pi to pi.

7. Aug 23, 2012

### Staff: Mentor

Why do you keep writing √(.36)? That simplifies to an exact value. What is this value?

I think you would be better off by NOT using cos-1, since that will give you only one value. I would sketch a graph of y = cos(2x) on the interval [$-2\pi, 2\pi$] (since x $\in$ [$-\pi, \pi$]), and identify all of the points at which cos(2x) = ±B, where B is the simplified value of √(.36).

EDIT: Also, your work above suggests that there are four solutions. I get quite a few more than that.

Last edited: Aug 23, 2012