+104 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 \(3\cot(4k-\pi)=\sqrt3\)
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.
+26367 Find all values of \(k\) such that \(3\cot(4k-\pi)=\sqrt3\)
\(\begin{array}{|rcll|} \hline \mathbf{3\cot(4k-\pi) } &=& \mathbf{ \sqrt3 } \quad &| \quad \cot(4k-\pi) = \dfrac{1}{\tan(4k-\pi)} \\\\ \dfrac{3}{\tan(4k-\pi)} &=& \sqrt3 \quad &| \quad \cdot \tan(4k-\pi) \\\\ 3 &=& \sqrt3 \cdot \tan(4k-\pi) \\\\ \sqrt3 \cdot \tan(4k-\pi) &=& 3 \quad &| \quad \cdot \sqrt3 \\ 3 \cdot \tan(4k-\pi) &=& 3\sqrt3 \quad &| \quad : 3 \\ \tan(4k-\pi) &=& \sqrt3 \\ \tan\Big( -(\pi-4k) \Big) &=& \sqrt3 \quad &| \quad \tan(-x)= - \tan(x) \\ -\tan(\pi-4k) &=& \sqrt3 \quad &| \quad \tan(\pi-x) = -\tan(x) \\ - \Big(-\tan(4k) \Big) &=& \sqrt3 \\ \mathbf{\tan(4k)} &=& \mathbf{\sqrt3} \quad & | \quad \arctan() \\ \arctan\Big(\tan(4k)\Big) &=& \arctan(\sqrt3) + n\cdot \pi \qquad n \in \mathbb{Z} \\ 4k &=& \arctan(\sqrt3) + n\cdot \pi \quad & | \quad \arctan(\sqrt3)=\dfrac{\pi}{3} \\ \\ 4k &=&\dfrac{\pi}{3}+ n\cdot \pi \quad & | \quad : 4 \\ \\ \mathbf{ k } &=& \mathbf{\dfrac{\pi}{12}+ n\cdot \dfrac{\pi}{4} } \qquad n \in \mathbb{Z} \\ \hline \end{array}\)
