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How to solve this trig equation

  • Thread starter lo2
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  • #1
lo2
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Homework Statement



Solve this equation:

[itex]cos^2(2x)=0,36[/itex]

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


Homework Equations



-

The Attempt at a Solution




[itex]cos^2(2x)=0,36 \Leftrightarrow cos(2x)=\sqrt{0,36} \Leftrightarrow 2x=cos^{-1}(\sqrt{0,36}) [/itex]

And then I am not sure exactly how to proceed... When should I put in the [itex] 2p \pi [/itex] where [itex] x \in Z [/itex], to get all of the possible solutions?
 
Last edited:

Answers and Replies

  • #2
33,521
5,199

Homework Statement



Solve this equation:

[itex]cos^2(2x)=0,36[/itex]

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


Homework Equations



-

The Attempt at a Solution




[itex]cos^2(2x)=0,36 \Leftrightarrow cos^2(2x)=\sqrt{0,36} \Leftrightarrow 2x=cos^{-1}(\sqrt{0,36}) [/itex]
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 [itex] 2p \pi [/itex] where [itex] x \in Z [/itex], to get all of the possible solutions?
 
  • #3
lo2
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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?
 
  • #4
SammyS
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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 ± .
 
  • #5
33,521
5,199
When should I put in the [itex] 2p \pi [/itex] where [itex] x \in Z [/itex], 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.
 
  • #6
lo2
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Ok I have come up with this solution:

[itex]\frac{cos^{-1}(\pm \sqrt{0,36})}{2}+p\pi[/itex]

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

Since the solutions have to be in the interval of -pi to pi.
 
  • #7
33,521
5,199
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:

[itex]\frac{cos^{-1}(\pm \sqrt{0,36})}{2}+p\pi[/itex]

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

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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