# `f(theta) = 2cos(theta) + sin^2(theta)` Find the critical numbers of the function

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You need to find the critical points of the function, hence, you need to evaluate the solutions to the equation `f'(theta) = 0` .

You need to evaluate the first derivative:

`f'(theta) = -2 sin theta + 2sin theta*cos theta`

You need to solve for theta `f'(theta) = 0` , such that:

`-2 sin theta + 2sin theta*cos theta= 0 `

Factoring out `2sin theta` yields:

`2sin theta*(-1 + cos theta) = 0`

Put `2sin theta = 0 => sin theta = 0`

`theta = (-1)^k*arcsin 0 + kpi`

`theta = kpi`

Put `-1 + cos theta= 0 =>cos theta = 1`

`theta = +-arccos 1 + 2kpi`

`theta = 2kpi`

**Hence, evaluating the critical numbers of the function for `f'(theta) = 0` , yields `theta =2kpi, theta = kpi.` **

`f(x) = 2*cos(x) + sin(x)^2 `

`f'(x) = 2sin(x) * (cos(x) - 1) `

To find the zeros:

`0 = -2sin(x) + 2sin(x)*cos(x) `

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

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

Therefore:

`sin(x) = 0 `

`x = arcsin(0) = 0 or pi or 2pi `

OR

`cos(x) -1 = 0 `

`cos(x) = 1 `

`x = arccos(1) `

`x = 0 or 2pi `