How do you find the exact solutions of the equation #tan2x-cotx=0# in the interval #[0,2pi)#?

2 Answers
Jan 23, 2017

#pi/6, pi/2#

Explanation:

tan 2x - cot x = 0
#(sin 2x)/(cos 2x) - cos x/(sin x) = 0#
#(sin x.sin 2x - cos x.cos 2x)/(sin x.cos 2x) = 0#
(sin x.sin 2x - cos x.cos 2x) = 0
Use trig identity: cos (a + b) = cos a.cos b - sin a.sin b
In this case we have:
sin x.cos 2x - cos x.cos 2x = - cos (x + 2x) = - cos 3x = 0
cos 3x = 0
Trig unit circle gives:
a. #3x = pi/2# --> #x = pi/6#
b. #3x = (3pi)/2# --> #x = (3pi)/6 = pi/2#
Answers for #(0, 2pi)#: #pi/6 and pi/2#
Check:
If #x = pi/6# --> #2x = pi/3# --> #tan 2x = sqrt3# --> #cot x = cot (pi/6) = sqrt3#, then: #tan 2x - cot x = sqrt3 - sqrt3 = 0#. OK
If #x = pi/2# --> #2x = pi# --> #tan 2x = tan pi = 0# --> #cot x = cot (pi/2) = 0#, then: #tan 2x - cot x = 0 - 0 = 0#. OK

Jan 23, 2017

Possible solutions in the interval #[0,2pi)# are #{pi/6,(5pi)/6,(7pi)/6,(11pi)/6}#

Explanation:

#tan2x-cotx=0#

#hArr(2tanx)/(1-tan^2x)=1/tanx#

or #2tan^2x=1-tan^2x#

or #3tan^2x-1=0#

i.e. #(sqrt3tanx+1)(sqrt3tanx-1)=0#

i.e. #tanx=1/sqrt3# or #-1/sqrt3#

Hence #x=npi+-pi/6#

and possible solutions in the interval #[0,2pi)# are

#{pi/6,(5pi)/6,(7pi)/6,(11pi)/6}#