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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
avatar
+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
avatar+23317 
+4

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

 

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

 

laugh

 Sep 20, 2019

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