Question

How to solve
cosx = sin(2x) on [0,2pi]

Answer 1

so what is the question?

Answer 2

It's part of a calculus question:
Find the relative extrema of f(x) = 2sin(x)+cos(2x) on [0,2pi] using the second derivative test.

Answer 3

cosx=2sinxcosx
=>cosx(1-2sinx)=0
=>cosx=0 or 1/2=sinx
so ans will be pi/2 ,3pi/2 ,5pi/6,pi/6

Answer 4

I have derived the function to find the critical values:
First derivative:
F(x) = cos(x)-2sin(2x)
Set to 0 and solve.... And we're back to my original question

Answer 5

2cos(x)-2sin(2x)**

Answer 6

Ahhh double angle formula! Thanks

Answer 7

How did you get cos(1-2sinx)=0 though??

Answer 8

sin(2x) = 2sin(x)cos(x)

so you will have cos(x) = 2cos(x)sin(x)

then rewrite the equation equal to 0 and solve by fractoring