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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 $\theta$?

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You do realise that $\tan(\theta)$ and $\sec(\theta)$ are undefined for $\theta=\frac{\pi}{2}+k\pi\,\,,\,k\in\Bbb Z$ – Henry Nov 10 '12 at 1:17

2 Answers 2

up vote 3 down vote accepted

$$8-2\sec^2\theta =0\Longrightarrow\sec^2\theta=4\Longrightarrow \cos^2\theta=\frac{1}{4}\Longrightarrow $$

$$\cos\theta=\pm\frac{1}{2}\Longrightarrow \theta=\pm\frac{\pi}{3}+k\pi\,\,,\,k\in\Bbb Z$$

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$0 = 8 - 2\sec^2\theta$
$2\sec^2\theta = 8$
$\sec^2\theta = 4$
$\sec\theta = \pm2$
$\cos\theta = \pm\frac{1}{2}$
$\theta = \frac{\pi}{3} + n\pi, \frac{2\pi}{3} + n\pi$, where $n\in\mathbb{Z}$

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