Question Number 148707 by mathdanisur last updated on 30/Jul/21
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](Q148722.png)
Commented bymathdanisur last updated on 30/Jul/21
Question Number 148707 by mathdanisur last updated on 30/Jul/21 | ||
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Answered by nimnim last updated on 30/Jul/21 | ||
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Commented bymathdanisur last updated on 30/Jul/21 | ||
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