Answer:
[tex]\theta_{1} = \frac{\pi}{3} \pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]
[tex]\theta_{2} = \frac{5\pi}{3} \pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]
Step-by-step explanation:
The critical numbers are found by the First Derivative Test, which consists in differentiating the function, equalizing it to zero and solving it:
[tex]g'(\theta) = 16 - 4\cdot \sec^{2} \theta[/tex]
Following equation needs to be solved:
[tex]16 - 4\cdot \sec^{2}\theta = 0[/tex]
[tex]\sec^{2}\theta = 4[/tex]
[tex]\cos^{2}\theta = \frac{1}{4}[/tex]
[tex]\cos \theta = \frac{1}{2}[/tex]
The solution is:
[tex]\theta = \cos^{-1} \frac{1}{2}[/tex]
Given that cosine is a periodical function, there are two subsets of solution:
[tex]\theta_{1} = \frac{\pi}{3} \pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]
[tex]\theta_{2} = \frac{5\pi}{3} \pm 2\pi\cdot i[/tex], [tex]\forall i \in \mathbb{N}_{O}[/tex]
