Find the critical values of the function: $g(\theta) = 8\theta - 2\tan(\theta)$
I found$\space g'(\theta)$: $g'(\theta)=8-2\sec^2(\theta)$
Then I set $g'(\theta) = 0$: $0 = 8-2\sec^2(\theta)$
Now how do I solve for $\theta$?
Find the critical values of the function: $g(\theta) = 8\theta - 2\tan(\theta)$
I found$\space g'(\theta)$: $g'(\theta)=8-2\sec^2(\theta)$
Then I set $g'(\theta) = 0$: $0 = 8-2\sec^2(\theta)$
Now how do I solve for $\theta$?
$8-2\sec^2\theta =0\Longrightarrow\sec^2\theta=4\Longrightarrow \cos^2\theta=\frac{1}{4}\Longrightarrow $
$\cos\theta=\pm\frac{1}{2}\Longrightarrow \theta=\pm\frac{\pi}{3}+k\pi\,\,,\,k\in\Bbb Z$
$0 = 8 - 2\sec^2\theta$
$2\sec^2\theta = 8$
$\sec^2\theta = 4$
$\sec\theta = \pm2$
$\cos\theta = \pm\frac{1}{2}$
$\theta = \frac{\pi}{3} + n\pi, \frac{2\pi}{3} + n\pi$, where $n\in\mathbb{Z}$