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anonymous
 5 years ago
dy/dx = 2sinx + 2cos2x
How do I find its critical points?
anonymous
 5 years ago
dy/dx = 2sinx + 2cos2x How do I find its critical points?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes I know this step, its just that I get a cos on one side and a sin on the other side. I don't know how to switch them to the same trig function to solve for x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0double angle cos(2x) = 1  2sin(x)^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so the I did the double angle and that gives me: 2sinx = 2(cos^2x  sin^2x) How did you get 1 2sin(x)^2?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cos(2x) = 1  2sin(x)^2 it's a rule you should have 2sinx = 2  2sin^2x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's called the double angle formula

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Funny, my book says: Cos2A = cos^2A  sin^2A
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