SOLUTION: Please help me solve this question: Find all solutions for 2/3Sin^2-cosx=1 on the interval (0,2pi)

Question 1056628: Please help me solve this question: Find all solutions for 2/3Sin^2-cosx=1 on the interval (0,2pi)
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Please help me solve this question: Find all solutions for 2/3Sin^2-cosx=1 on the interval (0,2pi)
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What do you need help with?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find all solutions for (2/3)Sin^2-cosx=1 on the interval (0,2pi)
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(2/3)(1-cos^2(x)) - cos(x) = 1
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(2/3) - (2/3)cos^2(x) - cos(x) = 1
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2 - 2cos^2(x) - 3cos(x) = 3
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2cos^2(x) + 3cos(x) + 1 = 0
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(2cos(x) + 1)(cos(x)+1) = 0
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cos(x) = -1/2 Or cos(x) = -1
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x = (2/3)pi or x = (4/3)pi or x = pi
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Cheers,
Stan H.
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