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twirlere
f(x)=2secx+tanx 0<x<2pi I'm finding critical points.
You mean where 1º derivative is 0?
? 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.
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