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# Trig Help

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Hi, I have this problem on my math book, I tried isolating k, but I'm not sure how to take it out of the cot(). I know Cot = 1/tan(x) but I think i'm missing something here:

The Problem is:

Find all values of $$k$$ such that $$3\cot(4k-\pi)=\sqrt3$$

Sep 20, 2019

#1
+1

Let 4k-$$\Pi$$$$\omega$$

So:

3 cot($$\omega$$)=$$\sqrt{3}$$

Divide both sides by 3

cot($$\omega$$)=$$\sqrt{3}/3$$

$$\omega$$ =$$cot^-1$$($$\sqrt{3}/3$$)

w=$$\Pi /3$$

so:

4k-$$\Pi$$=$$\Pi$$/3

k=1.04719755

approx: 1.05

pi/3

If your calculator doesn't have cot^-1 you can change it to 1/tan and process :)!

Hope that helped.

Also make sure what quadrant it is in.

Sep 20, 2019
#2
+23866
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Find all values of $$k$$ such that $$3\cot(4k-\pi)=\sqrt3$$

$$\begin{array}{|rcll|} \hline \mathbf{3\cot(4k-\pi) } &=& \mathbf{ \sqrt3 } \quad &| \quad \cot(4k-\pi) = \dfrac{1}{\tan(4k-\pi)} \\\\ \dfrac{3}{\tan(4k-\pi)} &=& \sqrt3 \quad &| \quad \cdot \tan(4k-\pi) \\\\ 3 &=& \sqrt3 \cdot \tan(4k-\pi) \\\\ \sqrt3 \cdot \tan(4k-\pi) &=& 3 \quad &| \quad \cdot \sqrt3 \\ 3 \cdot \tan(4k-\pi) &=& 3\sqrt3 \quad &| \quad : 3 \\ \tan(4k-\pi) &=& \sqrt3 \\ \tan\Big( -(\pi-4k) \Big) &=& \sqrt3 \quad &| \quad \tan(-x)= - \tan(x) \\ -\tan(\pi-4k) &=& \sqrt3 \quad &| \quad \tan(\pi-x) = -\tan(x) \\ - \Big(-\tan(4k) \Big) &=& \sqrt3 \\ \mathbf{\tan(4k)} &=& \mathbf{\sqrt3} \quad & | \quad \arctan() \\ \arctan\Big(\tan(4k)\Big) &=& \arctan(\sqrt3) + n\cdot \pi \qquad n \in \mathbb{Z} \\ 4k &=& \arctan(\sqrt3) + n\cdot \pi \quad & | \quad \arctan(\sqrt3)=\dfrac{\pi}{3} \\ \\ 4k &=&\dfrac{\pi}{3}+ n\cdot \pi \quad & | \quad : 4 \\ \\ \mathbf{ k } &=& \mathbf{\dfrac{\pi}{12}+ n\cdot \dfrac{\pi}{4} } \qquad n \in \mathbb{Z} \\ \hline \end{array}$$

Sep 20, 2019