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x-3-3-4-x-1-8-0-x-




Question Number 176995 by blackmamba last updated on 29/Sep/22
  x^3 −(3/4)x+(1/8)=0   x=?
$$\:\:{x}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{4}}{x}+\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$$\:{x}=? \\ $$
Answered by mr W last updated on 29/Sep/22
Δ=((1/(16)))^2 +(−(1/4))^3 =−(3/(256))<0  ⇒three real roots!  x=2(√(1/4)) sin ((1/3)sin^(−1) ((1/(16))/((1/4)(√(1/4))))+((2kπ)/3))  x=sin ((1/3)sin^(−1) (1/2)+((2kπ)/3))  x=sin ((π/(18))+((2kπ)/3)), k=0,1,2  x=sin 10^° , sin 130°, sin 250°
$$\Delta=\left(\frac{\mathrm{1}}{\mathrm{16}}\right)^{\mathrm{2}} +\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{3}} =−\frac{\mathrm{3}}{\mathrm{256}}<\mathrm{0} \\ $$$$\Rightarrow{three}\:{real}\:{roots}! \\ $$$${x}=\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{4}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\frac{\mathrm{1}}{\mathrm{16}}}{\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\frac{\mathrm{1}}{\mathrm{4}}}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right) \\ $$$${x}=\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right) \\ $$$${x}=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{18}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right),\:{k}=\mathrm{0},\mathrm{1},\mathrm{2} \\ $$$${x}=\mathrm{sin}\:\mathrm{10}^{°} ,\:\mathrm{sin}\:\mathrm{130}°,\:\mathrm{sin}\:\mathrm{250}° \\ $$
Commented by mr W last updated on 29/Sep/22
see Q155945
$${see}\:{Q}\mathrm{155945} \\ $$
Commented by Tawa11 last updated on 02/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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