Answer:
[tex]\theta = \cos^{-1} (\pm \frac{1}{2} )[/tex]
Step-by-step explanation:
The function is [tex]g(\theta) = 8\cdot \theta - 2\cdot \tan \theta[/tex], whose first derivative is:
[tex]g'(\theta) = 8 - 2\cdot \sec^{2}\theta[/tex]
Let equalize the derivative to zero:
[tex]8 - 2\cdot \sec^{2}\theta=0[/tex]
[tex]\sec^{2}\theta = 4[/tex]
[tex]1 = 4\cdot \cos^{2}\theta[/tex]
[tex]\cos \theta = \pm\sqrt {\frac{1}{4} }[/tex]
[tex]\cos \theta = \pm \frac{1}{2}[/tex]
The solutions are given by the following inverse trigonometric function:
[tex]\theta = \cos^{-1} (\pm \frac{1}{2} )[/tex]
