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

## Expert Answers 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.`

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