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

2 Answers
Jul 11, 2018

Answer:

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

Explanation:

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:

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

Explanation:

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#