How do you solve 3 tan 2 x = √ 3 tan x in interval 0<=x<=360?

Question

2446 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-02T05:03:23

#x={0^@,180^@,360^@}#

As #tan2x=(2tanx)/(1-tan^2x)#, we can write #3tan2x=sqrt3tanx# as

#3(2tanx)/(1-tan^2x)=sqrt3tanx#

or #6tanx=sqrt3tanx-sqrt3tan^3x#

or #sqrt3tan^3x+tanx(6-sqrt3)=0#

or #sqrt3tanx(tan^2x+2sqrt3-1)=0#

As #tan^2x+2sqrt3-1!=0#,

hence #tanx=0# i.e. #x=npi#, where #n# is an integer

and in #0^@<=x<=360^@#, #x={0^@,180^@,360^@}#.

Answer 2 · 2018-05-02T05:14:03

#=>x=0^circ,180^circ,360^circ.#

Here,

#3 tan 2 x = √ 3 tan x#

#=>3((2tanx)/(1-tan^2x))=sqrt3tanx#

#=>6tanx=sqrt3tanx-sqrt3tan^3x#

#=>sqrt3tan^3x+6tanx-sqrt3tanx=0#

#=>tanx(sqrt3tan^2x+6-sqrt3)=0#

#=>tanx=0 or sqrt3tan^2x+6-sqrt3=0#

#(i)tanx=0=>x=kpi,kinZZ#

#(ii)sqrt3tan^2x+6-sqrt3=0#

#=>tan^2x=-(6-sqrt3)/sqrt3=-(2sqrt3-1) <0#

#i.e.tan^2x <0=>x!inRR#

Hence,

#x=kpi,kinZZ ,where,x in [0^circ,360^circ]#

#=>x=0^circ,180^circ,360^circ.#