Question

f(x)=cotx
For what values of x is f(x) > -√3 on the interval (0,π)?

Answer 1

well cot = 1/tan

and tan is negative in the 2nd quadrant... so cot will be negative as well

so

\[\cot(x) = -\sqrt{3} ... then 1/\cot = \tan... so... \tan(x) = -\frac{1}{\sqrt3}\]

find the value of

\[x = \tan^{-1}(\frac{1}{\sqrt{3}})\]

and 2nd quadrant angle will be

\[\pi - x \]

Answer 2

Thank you :)