Respuesta :
Use formula
[tex]e^{ix}=\cos{x}+\sin{x}[/tex]
[tex]\sqrt[3]{27(\cos{330^o}+i\sin{330^o})} \\\sqrt[3]{27e^{i330^o} } \\ \sqrt[3]{27} \sqrt[3]{e^{i330^o}} \\ \sqrt[3]{3^3} \sqrt[3]{e^{i110^o\times 3}} \\ \sqrt[3]{3^3} \sqrt[3]{(e^{i110^o})^ 3}} \\ 3e^{i110^o} \\3(\cos{110^o}+i\sin{110^o})[/tex]
Since [tex]330^o=360^o+330^o=690^o [/tex] and [tex]330^o=2\times 360^o+330^o=1050^o [/tex] there are two more solutions:
1. [tex]3(\cos{230^o}+i\sin{230^o})[/tex]
2. [tex]3(\cos{350^o}+i\sin{350^o})[/tex]
[tex]e^{ix}=\cos{x}+\sin{x}[/tex]
[tex]\sqrt[3]{27(\cos{330^o}+i\sin{330^o})} \\\sqrt[3]{27e^{i330^o} } \\ \sqrt[3]{27} \sqrt[3]{e^{i330^o}} \\ \sqrt[3]{3^3} \sqrt[3]{e^{i110^o\times 3}} \\ \sqrt[3]{3^3} \sqrt[3]{(e^{i110^o})^ 3}} \\ 3e^{i110^o} \\3(\cos{110^o}+i\sin{110^o})[/tex]
Since [tex]330^o=360^o+330^o=690^o [/tex] and [tex]330^o=2\times 360^o+330^o=1050^o [/tex] there are two more solutions:
1. [tex]3(\cos{230^o}+i\sin{230^o})[/tex]
2. [tex]3(\cos{350^o}+i\sin{350^o})[/tex]
Answer:
The cube roots are
[tex]3(cos{110^{\circ}}+isin{110^{\circ}})[/tex]
[tex]3(cos{230^{\circ}}+isin{230^{\circ}})[/tex]
[tex]3(cos{350^{\circ}}+isin{350^{\circ}})[/tex]
Step-by-step explanation:
Given the expression 27(cos 330° + i sin 330°)
we have to find the cube roots of the above expression.
By Euler's formula,
[tex]e^{i\theta}=cos{\theta}+isin{\theta[/tex]
We can write
[tex]cos{330^{\circ}}+isin{330^{\circ}}=e^{330i}[/tex]
[tex]\sqrt[3]{27(cos{330^{\circ}}+isin{330^{\circ}})}\\\\=\sqrt[3]{3^3(cos{330^{\circ}}+isin{330^{\circ}})}\\\\=\sqrt[3]{3^3(e^{330i})}\\\\=\sqrt[3]{3^3(e^{110i\times3})}\\\\=\sqrt[3]{(3e^{110i})^3}\\\\=3e^{110i}=3(cos{110^{\circ}}+isin{110^{\circ}})[/tex]
Since, 330°=360°+330°=690° and 330°=2(360°)+330°=1050°
Other two solutions are
[tex]3(cos{230^{\circ}}+isin{230^{\circ}})[/tex] and [tex]3(cos{350^{\circ}}+isin{350^{\circ}})[/tex]
