How do you solve #tan x + tan 2x +tan3x=0#?

1 Answer
Nghi N
Jul 24, 2017

#x = (kpi)/3# or #k60^@#
#x = +- 35^@26 + k180^@#

Explanation:

f(x) = tan x + tan 2x + tan 3x = 0
First, apply trig identity:
#tan a + tan b = (tan a + tan b)/(1 - tan a.tan b)#
#tan x + tan 2x = (tan 3x)/(1 - tan x.tan 2x)#
#f(x) = (tan 3x)/(1 - tan x.tan 2x) + tan 3x#
Put tan 3x in common factor:
#f(x) = tan 3x(1/(1 - tan x.tan 2x) + 1) = 0#
Either factor must be zero.
a. tan 3x - 0 --> #3x = kpi# --> #x = (kpi)/3#
b. (1/(1 - tan x.tan 2x) + 1) = 0
1 + (1 - tan x.tan 2x) = 0
2 - tan x.tan 2x = 0
tan x.tan 2x = 2
#tan x ((2tan x)/(1 - tan^2 x)) = 2#
#2tan^2 x = 2 - 2tan^2 x#
#4tan^2 x = 2# --> #tan^2 x = 1/2# --> #tan x = +- sqrt2/2#
Calculator gives:
#x = +- 35^@26 + k180^@#
Check by calculator.
x = 60 --> #tan 60 = sqrt3# --> #tan 120 = - sqrt3# --> tan 180 = 0
tan 60 + tan 120 + tan 180 =# -sqrt3 + sqrt3 + 0 = 0#. Proved
x = 35.26 --> tan x = 0.71 --> tan 2x = 2.83 --> tan 3x = - 3.54
tan x + tan 2x + tan 3x = 0.71 + 2.83 - 3.54 = 0. Proved

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