Find any critical numbers of: f(theta) = 2sec(theta) + tan(theta), 0<theta<2pi.
I found the defivative, but I got stuck.
I got to 2tan(theta)=-sin(theta).
Oh, sorry. I got to include the steps.
f'(theta) = 2sec(theta)tan(theta) + sec^2(theta)
I set it equal to zero, factored out a sec(theta). I was left with 2 equations: sec(theta)=0 and 2tan(theta) + sec(theta) = 0 =>2tan(theta) = -sec(theta) {I didn't mean sin before...I meant sec}
ok. and you are aware that $\displaystyle \sec \theta $ is never zero, correct?
anyway, here's how i would proceed.
$\displaystyle 2 \tan \theta + \sec \theta = 0$
$\displaystyle \Rightarrow \frac {2 \sin \theta }{\cos \theta} + \frac 1{\cos \theta } = 0$
now combine the fractions and set the numerator equal to zero and solve for theta