# `h(x) = sin^2 (x) + cos(x), 0 < x < 2pi` Find the critical numbers of the function.

Given the function `h(x)=sin^2(x)+cos(x)` in the interval `0<x<2pi`

We have to find the critical numbers of the function.

First take the derivative of the function and equate it to zero.

We get,

`h'(x)=2sin(x)cos(x)-sin(x)=0`

`sin(x)(2cos(x)-1)=0`

`sin(x)=0`   or  `2cos(x)-1=0`

sin(x)=0 implies x= npi

i.e we get x= pi in the interval 0<x<2pi

...

Given the function `h(x)=sin^2(x)+cos(x)` in the interval `0<x<2pi`

We have to find the critical numbers of the function.

First take the derivative of the function and equate it to zero.

We get,

`h'(x)=2sin(x)cos(x)-sin(x)=0`

`sin(x)(2cos(x)-1)=0`

`sin(x)=0`   or  `2cos(x)-1=0`

sin(x)=0 implies x= npi

i.e we get x= pi in the interval 0<x<2pi

Now,

2cos(x)-1=0 implies cos(x)=1/2

So x= pi/3 and 5pi/3 (in the interval 0<x<2pi)

Hence the critical points are `x=pi/3, pi and (5pi)/3`

Approved by eNotes Editorial Team