Question Number 148707 by mathdanisur last updated on 30/Jul/21

Solve for equation:  2tg(3x) - 3tg(2x) = tg^2 (2x) ∙ tg(3x)

Answered by nimnim last updated on 30/Jul/21

Let me give a try....  ⇒2(((3tanx−tan^3 x)/(1−3tan^2 x)))−3(((2tanx)/(1−tan^2 x)))=(((2tanx)/(1−tan^2 x)))^2 (((3tanx−tan^3 x)/(1−3tan^2 x)))  ⇒((6tanx−6tan^3 x−2tan^3 x+2tan^5 x−6tanx+18tan^3 x)/((1−3tan^2 x)(1−tan^2 x)))=((4tan^2 x(3tanx−tan^3 x))/((1−tan^2 x)^2 (1−3tan^2 )))  ⇒((2tan^5 x+10tan^3 x)/1)=((4tan^3 x(3−tan^2 x))/(1−tan^2 x))  ⇒2tan^3 x(tan^2 x+5)(1−tan^2 x)−4tan^3 x(3−tan^2 x)=0  ⇒2tan^3 =0 or (tan^2 x+5)(1−tan^2 x)−6+2tan^2 x=0  ⇒tanx=0 or tan^2 x−tan^4 x+5−5tan^2 x−6+2tan^2 x=0  ⇒x=nπ, n∈Z   or    tan^4 x+2tan^2 x+1=0                                      or    (tan^2 x+1)^2 =0                                      or     tan^2 x+1=0                                      or     sec^2 x=0                                      or     secx=0  but the range of secx is (−∞,−1]∪[1,∞) and  zero does not fall within the range.           ∴ there is no solution.  Hence the only solution is x=nπ, n∈Z

Commented bymathdanisur last updated on 30/Jul/21

Thank you Ser