## anonymous 3 years ago f(x)=2secx+tanx 0<x<2pi I'm finding critical points.

1. anonymous

You mean where 1º derivative is 0?

2. anonymous

? I don't understand. I took the derivative and got: f'(x)=2secxtanx+secx2. Then I took out secx: secx(2tanx+secx). And I know I have to set it equal to 0 and solve, I'm having trouble with inside the parentheses though.

3. anonymous

write it like this: $\frac{1}{cosx}(\frac{2sinx}{cosx}+\frac{1}{cosx})=0$ now muñtiply bouth sides by cosx to get: 1(2sinx+1)=0 and solve for x

4. anonymous

ok?

5. anonymous

Okay! thanks!