# How to solve this trig equation

## Homework Statement

Solve this equation:

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

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

-

## 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?

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Mark44
Mentor

## Homework Statement

Solve this equation:

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

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

-

## The Attempt at a Solution

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

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?

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).
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?

SammyS
Staff Emeritus
Homework Helper
Gold Member
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?
Yes, use the ± .

Mark44
Mentor
When should I put in the $2p \pi$ where $x \in Z$, to get all of the possible solutions?
The domain for x is restricted to [##-\pi, \pi##], so you're going to get only a handful of solutions.

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.

Mark44
Mentor
Also, you should simplify √(.36).
Why do you keep writing √(.36)? That simplifies to an exact value. What is this value?

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.
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.

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