SOLUTION: solve exactly over 0 ° &#8804; &#952; < 360° : 4cos^2&#952; - 4sin&#952; =5

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Question 1057643: solve exactly over 0 ° ≤ θ < 360° : 4cos^2θ - 4sinθ =5
Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
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solve exactly over 0 ° ≤ θ < 360° : 4cos^2θ - 4sinθ =5
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4cos%5E2%28theta%29+-+4%2Asin%28theta%29 = 5.

Replace cos%5E2%28theta%29 by 1-sin%5E2%28theta%29 to make the equation uniform for sin%28theta%29. You will get

4%2A%281-sin%5E2%28theta%29%29+-+4%2Asin%28theta%29 = 5,   or

4%2Asin%5E2%28theta%29+%2B+4%2Asin%28theta%29+%2B+1 = 0,   or

%282%2Asin%28theta%29+%2B+1%29%5E2 = 0.


Then

2%2Asin%28theta%29+%2B+1 = 0  --->  sin%28theta%29 = -1%2F2  --->  theta = 210°  OR  theta = 330°.

Answer.  theta = 210°  OR  theta = 330°.