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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

$${a}\:{case}\:{with}\:{three}\:{real}\:{roots} \\ $$

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

$${we}\:{have} \\ $$$$\mathrm{sin}\:\mathrm{3}\theta=\mathrm{3}\:\mathrm{sin}\:\theta−\mathrm{4}\:\mathrm{sin}^{\mathrm{3}} \:\theta \\ $$$$\mathrm{sin}^{\mathrm{3}} \:\theta−\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{sin}\:\theta+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{3}\theta=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$$${let}\:{x}={a}\:\mathrm{sin}\:\theta,\:{a}\neq\mathrm{0} \\ $$$${a}^{\mathrm{3}} \mathrm{sin}^{\mathrm{3}} \:\theta−\mathrm{3}{a}\:\mathrm{sin}\:\theta+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin}^{\mathrm{3}} \:\theta−\frac{\mathrm{3}}{{a}^{\mathrm{2}} }\:\mathrm{sin}\:\theta+\frac{\mathrm{1}}{{a}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\Rightarrow−\frac{\mathrm{3}}{{a}^{\mathrm{2}} }=−\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow{a}=\mathrm{2} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}^{\mathrm{3}} }=\frac{\mathrm{sin}\:\mathrm{3}\theta}{\mathrm{4}}\:\Rightarrow\mathrm{sin}\:\mathrm{3}\theta=\frac{\mathrm{4}}{\mathrm{2}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{1}\:\checkmark \\ $$$$\Rightarrow\mathrm{3}\theta=\mathrm{2}{k}\pi+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow\theta=\frac{\mathrm{2}{k}\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{18}} \\ $$$$\Rightarrow{x}={a}\:\mathrm{sin}\:\theta=\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{18}}\right),\:{k}=\mathrm{0},\mathrm{1},\mathrm{2} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =\mathrm{2}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{18}}\right)\approx\mathrm{0}.\mathrm{347} \\ $$$$\Rightarrow{x}_{\mathrm{2}} =\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{18}}\right)=\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{13}\pi}{\mathrm{18}}\approx\mathrm{1}.\mathrm{532} \\ $$$$\Rightarrow{x}_{\mathrm{3}} =\mathrm{2}\:\mathrm{sin}\:\left(\frac{\mathrm{4}\pi}{\mathrm{3}}+\frac{\pi}{\mathrm{18}}\right)=\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{25}\pi}{\mathrm{18}}\approx−\mathrm{1}.\mathrm{878} \\ $$

Commented by Tawa11 last updated on 06/Oct/21

Great sir.

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

Commented by ArielVyny last updated on 06/Oct/21

sir can you solve x^3 −3x−3 ?

$${sir}\:{can}\:{you}\:{solve}\:{x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{3}\:? \\ $$

Commented by mr W last updated on 06/Oct/21

for case with one real root see Q155928

$${for}\:{case}\:{with}\:{one}\:{real}\:{root} \\ $$$${see}\:{Q}\mathrm{155928} \\ $$

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