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Question-155945




Question Number 155945 by mr W last updated on 06/Oct/21
Commented by mr W last updated on 06/Oct/21
a case with three real roots
acasewiththreerealroots
Answered by mr W last updated on 06/Oct/21
we have  sin 3θ=3 sin θ−4 sin^3  θ  sin^3  θ−(3/4) sin θ+(1/4)sin 3θ=0  x^3 −3x+1=0  let x=a sin θ, a≠0  a^3 sin^3  θ−3a sin θ+1=0  sin^3  θ−(3/a^2 ) sin θ+(1/a^3 )=0  ⇒−(3/a^2 )=−(3/4) ⇒a=2  ⇒(1/a^3 )=((sin 3θ)/4) ⇒sin 3θ=(4/2^3 )=(1/2)≤1 ✓  ⇒3θ=2kπ+sin^(−1) (1/2)=2kπ+(π/6)  ⇒θ=((2kπ)/3)+(π/(18))  ⇒x=a sin θ=2 sin (((2kπ)/3)+(π/(18))), k=0,1,2  ⇒x_1 =2 sin ((π/(18)))≈0.347  ⇒x_2 =2 sin (((2π)/3)+(π/(18)))=2 sin ((13π)/(18))≈1.532  ⇒x_3 =2 sin (((4π)/3)+(π/(18)))=2 sin ((25π)/(18))≈−1.878
wehavesin3θ=3sinθ4sin3θsin3θ34sinθ+14sin3θ=0x33x+1=0letx=asinθ,a0a3sin3θ3asinθ+1=0sin3θ3a2sinθ+1a3=03a2=34a=21a3=sin3θ4sin3θ=423=1213θ=2kπ+sin112=2kπ+π6θ=2kπ3+π18x=asinθ=2sin(2kπ3+π18),k=0,1,2x1=2sin(π18)0.347x2=2sin(2π3+π18)=2sin13π181.532x3=2sin(4π3+π18)=2sin25π181.878
Commented by Tawa11 last updated on 06/Oct/21
Great sir.
Greatsir.
Commented by ArielVyny last updated on 06/Oct/21
sir can you solve x^3 −3x−3 ?
sircanyousolvex33x3?
Commented by mr W last updated on 06/Oct/21
for case with one real root  see Q155928
forcasewithonerealrootseeQ155928

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