How do you solve #t a n x + \tan 2x + \tan 3x = 0#?

2 Answers
Nghi N
Jul 16, 2017

#x = kpi#
#x = +- 35^@18 + k180^@#
#x = +- 60^@ + k180^@#

Explanation:

Cal tan x = t and apply trig identities:
#tan 2a = (2tan a)/(1 - tan^2 a)#
#tan 3a = (3tan a - tan^3a)/(1 - 3tan^2 a)#
We get the equation in tan x = t -->
#f(t) = t + (2t)/(1 - t^2) + (3t - t^3)/(1 - 3t^2) = 0#
Put #(1 - t^2)(1 - 3t^2)# in common denominator, then, develop the numerator:
#t(1 - t^2)(1 - 3t^2) + 2t(1 - 3t^2) + (3t - t^3)(1 - t^2) = 0#
After development:
#f(t) = 2t(2t^4 - 7t^2 + 3) = 0#
Factor the trinomial in parentheses:
#f(t) = 2t(2t^2 - 1)(t^2 - 3) = 0#
Either factor must be zero.
a. tan x = t = 0 --> x = 0 --> and #x = pi#
b. #2t^2 - 1 = 0# --> #t^2 = 1/2# --> #t = +- sqrt2/2#
#tan x = t = sqrt2/2# --> #x = 35^@ 18 + k180^@#
#tan x = t = - sqrt2/2# --> #x = - 35^@18 + k180^@#
c. #t^2 = 3# --> #t = +- sqrt3#
#t = tan x = sqrt3# --> #x = 60^@ + k180^@#
#t = tan x = - sqrt3# --> #x = - 60^@ + k180^@#
Check by calculator.
x = 60 --> tan x = 1.732 --> 2x = 120 --> tan 2x = - 1.732 --> 3x = 180 --> tan 180 = 0.
tan 60 + tan 120 + tan 180 = 1.73 - 1.73 + 0 = 0 Proved
x = 35.18 --> tan x = 0.70 --> tan 2x = tan 70.36 = 2.80 -->
tan 3x = tan 105.54 = - 3.50.
tan x + tan 2x + tan 3x = 0.70 + 2.80 - 3.50 = 0. Proved.

P dilip_k
Jul 16, 2017

#tanx+tan2x+tan3x=0#

#(1-tanxtan2x)((tanx+tan2x)/(1-tanxtan2x))+tan3x=0#

#=>(1-tanxtan2x)tan3x+tan3x=0#

#=>tan3x(2-tanxtan2x)=0#

When #tan3x=0#

#=>3x=npi" where "nin ZZ#

#=>x=(npi)/3" where "nin ZZ#

when

#(2-tanxtan2x)=0#

#=>tanxtan2x=2#

#=>(2tan^2x)/(1-tan^2x)=2#

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

#=>2tan^2x=1#

#=>tanx=pm1/sqrt2=tan(pm0.2pi)#
#=>x=npipm 0.2pi" where "nin ZZ#

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