SOLUTION: Solve the equation on interval [0,2pi)? 8 cos^2 x -4 = 0 thanks in advance

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Question 919594: Solve the equation on interval [0,2pi)?

8 cos^2 x -4 = 0
thanks in advance
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
8%2Acos%5E2%28x%29+-+4+=+0

8%2Acos%5E2%28x%29+=+4

cos%5E2%28x%29+=+4%2F8

cos%5E2%28x%29+=+1%2F2

cos%28x%29+=+%22%22%2B-sqrt%281%2F2%29

cos%28x%29+=+sqrt%281%2F2%29 or cos%28x%29+=+-sqrt%281%2F2%29

cos%28x%29+=+sqrt%282%29%2F2 or cos%28x%29+=+-sqrt%282%29%2F2

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Solve cos%28x%29+=+sqrt%282%29%2F2 to get x+=+pi%2F4 or x+=+5pi%2F4

Use the unit circle to find these two angles.

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Solve cos%28x%29+=+-sqrt%282%29%2F2 to get x+=+3pi%2F4 or x+=+7pi%2F4

Use the unit circle to find these two angles.
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The four solutions in [0,2pi) are x+=+pi%2F4, x+=+5pi%2F4, x+=+3pi%2F4 or x+=+7pi%2F4


Let me know if you need more help or if you need me to explain a step in more detail.
Feel free to email me at jim_thompson5910@hotmail.com
or you can visit my website here: http://www.freewebs.com/jimthompson5910/home.html

Thanks,

Jim

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the equation on interval [0,2pi)?
8 cos^2 x -4 = 0
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cos^2(x) = 4/8 = 1/2
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cos(x) = 1/sqrt(2)
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x = pi/4 or x = (7/4)pi
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Cheers,
Stan H.
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