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Question Number 157489 by MathSh last updated on 23/Oct/21
Solve for real numbers: tan(10x) = tan^5 (2x)
Solveforrealnumbers:tan(10x)=tan5(2x)
Answered by MJS_new last updated on 24/Oct/21
let t=tan 2x ((t(t^4 −10t^2 +5))/(5t^4 −10t^2 +1))=t^5 (t+1)^3 t(t−1)^3 (t^2 +1)=0 t=−1∨t=0∨t=1 x=−(π/8)+((nπ)/2)∨x=((nπ)/2)∨x=(π/8)+((nπ)/2)
lett=tan2xt(t4−10t2+5)5t4−10t2+1=t5(t+1)3t(t−1)3(t2+1)=0t=−1∨t=0∨t=1x=−π8+nπ2∨x=nπ2∨x=π8+nπ2
Commented by tounghoungko last updated on 24/Oct/21
tan 5x=((tan^5 x−10tan^3 x+5tan x)/(1−10tan^2 x+5tan^4 x))
tan5x=tan5x−10tan3x+5tanx1−10tan2x+5tan4x
Commented by MJS_new last updated on 24/Oct/21
yes
yes
Commented by MathSh last updated on 24/Oct/21
Thank you dear Ser very nice
ThankyoudearServerynice
Answered by tounghoungko last updated on 24/Oct/21
Solve for x∈R . tan (10x)=tan^5 (2x) ⇔ tan (5θ)=((sin^5 θ)/(cos^5 θ)) ,[θ=2x] ⇔((sin 5θ)/(cos 5θ)) = ((sin^5 θ)/(cos^5 θ)) ⇔ ((5sin θ−20sin^3 θ+16sin^5 θ)/(5cos θ−20cos^3 θ+16cos^5 θ))=((sin^5 θ)/(cos^5 θ)) ⇔5sin θcos^5 θ−20sin^3 θcos^5 θ+16sin^5 θcos^5 θ= 5cos θsin^5 θ−20cos^3 θsin^5 θ+16cos^5 θsin^5 θ ⇔sin θcos θ [5cos^4 θ−20sin^2 θcos^4 θ+16sin^4 θcos^4 θ− [5sin^4 θ+20cos^2 θsin^4 θ−16sin^4 θcos^4 θ ]=0 ⇔ (1/2)sin 2θ [5(cos^4 θ−sin^4 θ)+20(cos^2 θsin^4 θ−sin^2 θcos^4 θ)]=0 ⇔(1/2)sin 2θ [5(cos^2 θ−sin^2 θ)+20cos^2 θsin^2 θ(sin^2 θ−cos^2 θ)]=0 ⇔(5/2)sin 2θ cos 2θ (1−sin^2 2θ)=0 ⇔(5/4)sin 4θ (1−sin^2 2θ)=0 (i) (5/4)sin 4θ=0⇒ { ((4θ=2nπ)),((4θ=(1+2n)π)) :} ⇒ { ((8x=2nπ)),((8x=(1+2n)π)) :}⇒ { ((x=((nπ)/4))),((x=(((1+2n)/8))π)) :} (ii) 1−sin^2 2θ=0 ⇒ { ((sin 2θ=1)),((sin 2θ=−1)) :} ⇒ { (( 2θ=(π/2)+2kπ)),((2θ=((3π)/2)+2kπ)) :} { ((4x=(π/2)+2kπ)),((4x=((3π)/2)+2kπ)) :}⇒ { ((x=(((4k+1)/8))π)),((x=(((3+4k)/8))π)) :}
Solveforx∈R.tan(10x)=tan5(2x)⇔tan(5θ)=sin5θcos5θ,[θ=2x]⇔sin5θcos5θ=sin5θcos5θ⇔5sinθ−20sin3θ+16sin5θ5cosθ−20cos3θ+16cos5θ=sin5θcos5θ⇔5sinθcos5θ−20sin3θcos5θ+16sin5θcos5θ=5cosθsin5θ−20cos3θsin5θ+16cos5θsin5θ⇔sinθcosθ[5cos4θ−20sin2θcos4θ+16sin4θcos4θ−[5sin4θ+20cos2θsin4θ−16sin4θcos4θ]=0⇔12sin2θ[5(cos4θ−sin4θ)+20(cos2θsin4θ−sin2θcos4θ)]=0⇔12sin2θ[5(cos2θ−sin2θ)+20cos2θsin2θ(sin2θ−cos2θ)]=0⇔52sin2θcos2θ(1−sin22θ)=0⇔54sin4θ(1−sin22θ)=0(i)54sin4θ=0⇒{4θ=2nπ4θ=(1+2n)π⇒{8x=2nπ8x=(1+2n)π⇒{x=nπ4x=(1+2n8)π(ii)1−sin22θ=0⇒{sin2θ=1sin2θ=−1⇒{2θ=π2+2kπ2θ=3π2+2kπ{4x=π2+2kπ4x=3π2+2kπ⇒{x=(4k+18)πx=(3+4k8)π
Commented by MathSh last updated on 24/Oct/21
Thank you dear Ser cool
ThankyoudearSercool

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