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Question Number 95495 by i jagooll last updated on 25/May/20

3cos^2 x − 3cos x sin x + 2sin x = 1 x ∈ [ 0, 2π ]

3cos2x−3cosxsinx+2sinx=1x∈[0,2π]

Answered by bobhans last updated on 25/May/20

3−3sin^2 x−3sin xcos x +2sin x = 1 sin x (3(√(1−sin^2 x)) −2)= 2−3sin^2 x let s = sin x ⇒ s (3(√(1−s^2 ))−2) = 2−3s^2 s^2 (9−9s^2 −6(√(1−s^2 )) +4 ) = 4−12s^2 +9s^4 s^2 (13−9s^2 −6(√(1−s^2 )) ) = 4−12s^2 +9s^4 ...very hard

3−3sin2x−3sinxcosx+2sinx=1sinx(31−sin2x−2)=2−3sin2xlets=sinx⇒s(31−s2−2)=2−3s2s2(9−9s2−61−s2+4)=4−12s2+9s4s2(13−9s2−61−s2)=4−12s2+9s4...veryhard

Answered by john santu last updated on 25/May/20

Commented by john santu last updated on 25/May/20

by desmos

bydesmos

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