How do you solve #sqrt3+tan(2x)=0# between the interval #0<=x<=2pi#?

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How do you solve #sqrt3+tan(2x)=0# between the interval #0<=x<=2pi#?

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Answer 1 · 2018-03-21T02:57:00

#pi/3; (5pi)/6#

Explanation:

#tan 2x = -sqrt3#
Trig table and unit circle gve 2 solutions:
#2x = (2pi/3)#, and #2x = (2pi/3) + pi = (5pi)/3#
a. #2x = (2pi)/3#
#x = pi/3#
b. #2x = (5pi)/3#
#x = (5pi)/6#
Check.
#x = pi/3 # --> #tan 2x = tan ((2pi)/3) = -sqrt3#.
#sqrt3 + tan 2x = sqrt3 - sqrt3 = 0#. Proved
#x = (5pi)/6# --> #tan (10pi)/6 = tan (5pi)/3 = tan (2pi)/3 = - sqrt3#
#sqrt3 + tan 2x = sqrt3 - sqrt3 = 0#. Proved.

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How do you solve #sqrt3+tan(2x)=0# between the interval #0<=x<=2pi#?

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