Can cot(x)=0 and what are the values.

And also, how would I solve this.

$\displaystyle cot^2x-cotx=0$

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- Mar 28th 2008, 09:40 PM #1

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- Mar 28th 2008, 09:52 PM #2

$\displaystyle \cot(x)=\frac{\cos(x)}{\sin(x)}$

so it has the same zero's as the cosine function

$\displaystyle \cot^2x-\cot(x) =0 \iff \cot(x) [\cot(x)-1]=0$

so $\displaystyle \cot(x) = 0$ or $\displaystyle \cot(x)=1$

$\displaystyle x=\pi \cdot k \mbox{ } k \in \mathbb{Z}$ and for the 2nd

$\displaystyle x=\frac{\pi}{4}+\pi \cdot k \mbox{ } k \in \mathbb{Z} $