Question

dy/dx = -2sinx + 2cos2x

How do I find its critical points?

Answer 1

set =0

Answer 2

Yes 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

Answer 3

double angle
cos(2x) = 1 - 2sin(x)^2

Answer 4

ok 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?

Answer 5

cos(2x) = 1 - 2sin(x)^2
it's a rule

you should have
2sinx = 2 - 2sin^2x

Answer 6

it's called the double angle formula

Answer 7

Funny, my book says:

Cos2A = cos^2A - sin^2A