Okay, I always struggle with how to exactly work out the algebra for these trigonometric problems... The question is to identify where and what type of critical points lie on the function:

$\displaystyle f(x)=cos(2x)-1sin(x)$

between 0 and 2pi. Apparently there are 4.

So far this is what I have done:

$\displaystyle f'(x)=-2sin(2x)-cosx$

$\displaystyle 0=-4sinxcosx-cosx$

$\displaystyle cosx=-4sinxcosx$

$\displaystyle -4sinx=1$

$\displaystyle sinx=-1/4$

First of all I don't know how to solve for x in the final line without a calculator. Also, I only see that that would only possibly lead to two critical points between 0 and 2pi....

Thank you anyone who can help me a bit!

P.S. I kind of suck at this math formatting so forgive me...