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

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Answer 1 · 2018-07-11T14:33:25

$pi/2,3*pi/2,3*pi/4,7*pi/4$

Factorizing your equation we get

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

$x=pi72$ or $x=3pi/2$
for the second case we get

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

$x=3pi/4$ or $x=7pi/4$

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

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

Answer 2 · 2018-07-11T14:41:40

$x=90^\circ, 135^\circ, 270^\circ, 315^\circ$

Given that

$\cot^2x+\cot x=0$

$\cot x(\cot x+1)=0$

$\cot x=0\ \ or\ \ \cot x=-1$

$\tan x=\infty\ \ or \ \ \tan x=-1$

$x=n\pi+\pi/2\ \ or \ \ x=n\pi-\pi/4$

Where, $n$ is any integer i.e. $n=0, \pm1, \pm2, \pm3, \ldots$

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

$x=\pi/2, {3\pi}/2\ \ or\ \ x={3\pi}/4, {7\pi}/4$

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

$x=90^\circ, 135^\circ, 270^\circ, 315^\circ$

How do you solve cot^2x+cotx=0cot^2x+cotx=0 and find all solutions in the interval 0<=x<3600<=x<360?