solving a trig equation

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r8654

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Apr 5, 2010
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hello, could anyone help me solve the problem of ?(2)cos(2x)=1
The square root is over the 2 if it is hard to read.
I know the answers is pi/8, 7pi/8, 9pi/8. and 15pi/8, but i do not know how to get there. My attempts at using the double angle of cosine never seem to work.

Any help would be appreciated.
 
r8654 said:
hello, could anyone help me solve the problem of ?(2)cos(2x)=1
The square root is over the 2 if it is hard to read.
I know the answers is pi/8, 7pi/8, 9pi/8. and 15pi/8, but i do not know how to get there. My attempts at using the double angle of cosine never seem to work.

Any help would be appreciated.
Click to expand...

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
 
Well my first issue is that i do not know how to get rid of the ?(2) and the double angle i tried using was cos^2-sin^2. Whatever I try i seem to get a cluster of cos and sin.
I think what i really need to know is what to do after I use the double angle.

Thank you for replying.
 
r8654 said:
Well my first issue is that i do not know how to get rid of the ?(2) and the double angle i tried using was cos^2-sin^2. Whatever I try i seem to get a cluster of cos and sin.
I think what i really need to know is what to do after I use the double angle.

Thank you for replying.
Click to expand...

\(\displaystyle cos(2n\pi \pm \frac{\pi}{4}) \ \ = \ \ \frac{1}{\sqrt{2}}\)
 
\(\displaystyle \sqrt2 cos(2x) \ = \ 1,\)

\(\displaystyle cos(2x) \ = \ \frac{1}{\sqrt 2}\)

\(\displaystyle 2x \ = \ arccos\bigg(\frac{1}{\sqrt 2}}\bigg) \ = \ \frac{\pi}{4}\)

\(\displaystyle Hence, \ x \ = \ \frac{\pi}{8}+k\pi, \ \frac{15 \pi}{8}+k\pi, \ k \ an \ integer.\)
 
thank you super easy now. did not even thing of using arccos there.
 
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