# How do you solve cot^2x+cotx=0 and find all solutions in the interval 0<=x<360?

Jul 11, 2018

$\frac{\pi}{2} , 3 \cdot \frac{\pi}{2} , 3 \cdot \frac{\pi}{4} , 7 \cdot \frac{\pi}{4}$

#### Explanation:

$\cot \left(x\right) = 0$ so $\cos \left(x\right) = 0$
we get the following Solutions in the given interval

$x = \pi 72$ or $x = 3 \frac{\pi}{2}$
for the second case we get

$\cot \left(x\right) = - 1$ so $\cos \left(x\right) = - \sin \left(x\right)$
and we get

$x = 3 \frac{\pi}{4}$ or $x = 7 \frac{\pi}{4}$

Hint:
To get the Solutions of the last equation you can use

sqrt(sin(x+pi/4)=0

$x = {90}^{\setminus} \circ , {135}^{\setminus} \circ , {270}^{\setminus} \circ , {315}^{\setminus} \circ$

#### Explanation:

Given that

$\setminus {\cot}^{2} x + \setminus \cot x = 0$

$\setminus \cot x \left(\setminus \cot x + 1\right) = 0$

$\setminus \cot x = 0 \setminus \setminus \mathmr{and} \setminus \setminus \setminus \cot x = - 1$

$\setminus \tan x = \setminus \infty \setminus \setminus \mathmr{and} \setminus \setminus \setminus \tan x = - 1$

$x = n \setminus \pi + \setminus \frac{\pi}{2} \setminus \setminus \mathmr{and} \setminus \setminus x = n \setminus \pi - \setminus \frac{\pi}{4}$

Where, $n$ is any integer i.e. $n = 0 , \setminus \pm 1 , \setminus \pm 2 , \setminus \pm 3 , \setminus \ldots$

But given that $x \setminus \in \left[0 , {360}^{\setminus} \circ\right]$, hence setting $n = 0 , 1$ in first general solution & $n = 1 , 2$ in second general solutions, we get

$x = \setminus \frac{\pi}{2} , \frac{3 \setminus \pi}{2} \setminus \setminus \mathmr{and} \setminus \setminus x = \frac{3 \setminus \pi}{4} , \frac{7 \setminus \pi}{4}$

$x = \setminus \frac{\pi}{2} , \frac{3 \setminus \pi}{4} , \frac{3 \setminus \pi}{2} , \frac{7 \setminus \pi}{4}$

$x = {90}^{\setminus} \circ , {135}^{\setminus} \circ , {270}^{\setminus} \circ , {315}^{\setminus} \circ$