Find the local maximum and local minimum values of `f(x)=secx+logcos^2x ,\ \ 0ltxlt2pi`

Explanation

`f(x)=sec x+2 log cos x`



Therefore, `f^{prime}(x)=sec x tan x-2 tan x=tan x(sec x-2)`



`f^{prime}(x)=0 `


`Rightarrow tan x=0 text { or } sec x=2 text { or } cos x=frac{1}{2}`



Therefore, possible values of `x` are `x=0`


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