How Does One Find All Solutions to cos²(2x) = 0.36 in the Interval [-π, π]?

[lo2]

Homework Statement

Solve this equation:

cos^2(2x)=0,36

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

Homework Equations

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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?

 

[Mark44]

Homework Statement

Solve this equation:

cos^2(2x)=0,36

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

Homework Equations

-

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?

 

[lo2]

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]

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]

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.

 

[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.

 

[Mark44]

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.