You need to remember that the function `f(theta)` has critical values if the equation `f'(theta)=0` has solutions, hence, you neded to evaluate `f'(theta)` such that:

`f'(theta) = (6sec theta + 3tan theta)'`

You should remember that `sec theta = 1/cos theta` , hence, you need to use quotient rule to evaluate (sec...

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You need to remember that the function `f(theta)` has critical values if the equation `f'(theta)=0` has solutions, hence, you neded to evaluate `f'(theta)` such that:

`f'(theta) = (6sec theta + 3tan theta)'`

You should remember that `sec theta = 1/cos theta` , hence, you need to use quotient rule to evaluate (sec theta)' such that:

`(sec theta)' = (1'*cos theta - 1*(cos theta)')/(cos^2 theta)`

`(sec theta)' = -(-sin theta)/(cos^2 theta)=> (sec theta)' = (sin theta)/(cos^2 theta)`

`f'(theta) = 6sin theta/(cos^2 theta) + 3/(cos^2 theta)`

Factoring out `3/(cos^2 theta)` yields:

`f'(theta) = 3/(cos^2 theta)(2sin theta + 1)`

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

`3/(cos^2 theta)(2sin theta + 1) = 0`

Since `3/(cos^2 theta)!=0 => 2sin theta + 1 = 0` such that:

`2sin theta + 1 = 0 => 2sin theta=- 1 => sin theta = -1/2`

You need to remember that the sine function is negative in quadrants 3 and 4, hence, you will have the following solutions to the equation `sin theta = -1/2` such that:

`sin theta = -1/2 => theta = pi+pi/6 => theta = 7pi/6` (quadrant 3)

`theta = 2pi-pi/6 => theta = 11pi/6` (quadrant 4)

**Hence, evaluating the critical values of the function yields `theta = 7pi/6` and `theta = 11pi/6` .**