Which of the following are solutions to the equation
cos^2 (2x) - 1/4 = 0?
Check ALL that apply.
A. 5 pi/ 6
B. 11pi / 3
C. pi / 6
D. 12pi / 6

- anonymous

- chestercat

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- anonymous

From doing some current calculations. I came to the conclusion of C. Not sure if it is right or if there are anymore answers

- SolomonZelman

\(\large\color{black}{ \displaystyle {\rm Cos}^2 (2x) - \frac{1}{4} = 0 }\)
\(\large\color{black}{ \displaystyle {\rm Cos}^2 (2x)= \frac{1}{4} }\)
\(\large\color{black}{ \displaystyle \left(~~ {\rm Cos} (2x)~\right)^2= \left(\frac{1}{2}\right)^2 }\)

- SolomonZelman

hope that helps...

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## More answers

- SolomonZelman

(take the square root of both sides, and don't forget the \(\pm\) )

- anonymous

Ok! I will try to solve from there and I'll tell you what I get

- SolomonZelman

sure:)

- SolomonZelman

(oh, there is only 1 incorrect option)

- anonymous

So \[\cos(2x) = \pm \frac{ 1 }{ 2 }\] If I remember how to do this properly, I think I find what 1/2 is on the unit circle, which would be 60?
Then divide 60 by 2 and get 30 degrees?

- anonymous

So there are 3 correct answers?

- SolomonZelman

yes

- SolomonZelman

30 degrees is one of them (as you said)

- anonymous

Ok! Now I do not know where to do from here

- anonymous

We already have \[\frac{ \pi }{ 6}\] as one answer

- SolomonZelman

you can plug the rest of the options, if you don't want to go ahead and generate the solution sets for -1/2 and +1/2.

- SolomonZelman

yes π/6 is correct

- anonymous

I don't think I plugged the numbers in properly, but I got 12pi/6 and 5pi/6?

- SolomonZelman

12π/6 is not right

- SolomonZelman

5π/6 is right

- anonymous

Ok! So how would I plug them in? I know I did it wrong haha

- SolomonZelman

(click alt, and hold it
click 2 2 7 respectively on your number pad on the left if you have one
release alt: you get π)

- SolomonZelman

you would just plug the answer choices instead of x, into that equation
cos²(2x)-1/4=0

- anonymous

Oh and what gets you = 0 right?

- anonymous

\[2 \cos ^22x=\frac{ 1 }{ 2 }\]
\[1+\cos 4x=\frac{ 1 }{ 2 }\]
\[\cos 4x=\frac{ 1 }{ 2 }-1=-\frac{ 1 }{ 2 }=-\cos \frac{ \pi }{ 3 }=\cos \left(2n \pi+ \pi \pm \frac{ \pi }{ 3 } \right)\]
where n is an integer.
\[4x=\left( 2n+1 \right)\pi \pm \frac{ \pi }{ 3 }\]
\[x=\frac{ 1 }{ 4 }\left\{ \left( 2n+1 \right)\pi \pm \frac{ \pi }{ 3 } \right\}\]
now you can check by plugging n=0,1,2,....

- SolomonZelman

cos(4x) :) I haven't thought of that one....

- anonymous

So the answers would be \[\frac{ 5\pi }{ 6}, \frac{ 11\pi }{ 3 }, and \frac{ \pi }{ 6 }?\]

- SolomonZelman

yes.
use ~ for space

- SolomonZelman

yw

- anonymous

Thank you for your time and the tips :)

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