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Question Number 136008 by liberty last updated on 17/Mar/21
cosxtanx=sin3x+cos3x
Answered by MJS_new last updated on 17/Mar/21
t=tanx∧t⩾0t(t4+1)1/2=t6+1(t4+1)3/2t6−t5−t+1=0(t−1)(t5−1)=0⇒t=1⇒x=π4+nπ
Answered by liberty last updated on 18/Mar/21
notethattanx⩾0⇒sinxcosx=(sinx+cosx)(1−sinxcosx)⇒sinxcosx=1+2sinxcosx(1−sinxcosx)letsinxcosx=ℓ⇒ℓ=1+2ℓ(1−ℓ)⇒ℓ=(1+2ℓ)(1−2ℓ+ℓ2)⇒ℓ=1−2ℓ+ℓ2+2ℓ−4ℓ2+2ℓ3⇒2ℓ3−3ℓ2−ℓ+1=0⇒(2ℓ−1)(ℓ2−ℓ−1)=0⇒{ℓ=12⇒sinxcosx=12,sin2x=1ℓ=1±52(rejected)sosolutionis2x=π2+2nπorx=π4+nπ
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