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Cos(2θ) = 5 sin(θ) − 2

User Wesly
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1 Answer

2 votes
Cos(2
cos2 \alpha = 1-sin^(2) \alpha.

so...


1-sin^2\alpha = 5sin\alpha -2 sin^2\alpha + 5sin\alpha -3 = 0

Is that what you are trying to do? Find the zeros?
User Jackyesind
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