find the critical numbers of the function. (enter your answers as a comma - separated list. use n to denote…

Explanation:

Step1: Differentiate the function

The derivative of (g(\theta)=16\theta - 4\tan(\theta)) is (g'(\theta)=16-4\sec^{2}(\theta)) (using the rules (\frac{d}{d\theta}(a\theta)=a) and (\frac{d}{d\theta}(\tan\theta)=\sec^{2}\theta)).

Step2: Set the derivative equal to zero

Set (g'(\theta) = 0), so (16-4\sec^{2}(\theta)=0). First, divide both sides by (4): (4-\sec^{2}(\theta)=0). Since (\sec^{2}\theta=\frac{1}{\cos^{2}\theta}), we have (4-\frac{1}{\cos^{2}\theta}=0). Then, (\frac{1}{\cos^{2}\theta}=4), so (\cos^{2}\theta=\frac{1}{4}), and (\cos\theta=\pm\frac{1}{2}).

Step3: Solve for (\theta)

When (\cos\theta=\frac{1}{2}), (\theta = 2n\pi\pm\frac{\pi}{3}) (where (n\in\mathbb{Z})). When (\cos\theta=-\frac{1}{2}), (\theta = 2n\pi\pm\frac{2\pi}{3}) (where (n\in\mathbb{Z})).

Answer:

(\theta = 2n\pi\pm\frac{\pi}{3},2n\pi\pm\frac{2\pi}{3})