English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometria >

sin^3(x)=-1

Solução

sin^{3} (x) = - 1

Solução

x = \frac{3 π}{2} + 2 πn

Graus

x = 27 0^{\circ} + 36 0^{\circ} n

Passos da solução

sin^{3} (x) = - 1

Usando o método de substituição

sin^{3} (x) = - 1

Sea:

sin (x) = u

u^{3} = - 1

u^{3} = - 1 : u = - 1, u = \frac{1}{2} - i \frac{3}{2}, u = \frac{1}{2} + i \frac{3}{2}

u^{3} = - 1

Para

x^{3} = f (a)

as soluções são

x = 3 f (a), 3 f (a) \frac{- 1 - 3 i}{2}, 3 f (a) \frac{- 1 + 3 i}{2}

u = 3 - 1, u = 3 - 1 \frac{- 1 + 3 i}{2}, u = 3 - 1 \frac{- 1 - 3 i}{2}

3 - 1 = - 1

3 - 1

n - 1 = - 1,

n

é ímpar

3 - 1 = - 1

= - 1

Simplificar

3 - 1 \frac{- 1 + 3 i}{2} : \frac{1}{2} - i \frac{3}{2}

3 - 1 \frac{- 1 + 3 i}{2}

3 - 1 = - 1

3 - 1

n - 1 = - 1,

n

é ímpar

3 - 1 = - 1

= - 1

= - 1 \cdot \frac{- 1 + 3 i}{2}

Multiplicar:

1 \cdot \frac{- 1 + 3 i}{2} = \frac{- 1 + 3 i}{2}

= - \frac{- 1 + 3 i}{2}

Reescrever

- \frac{- 1 + 3 i}{2}

na forma complexa padrão:

\frac{1}{2} - \frac{3}{2} i

- \frac{- 1 + 3 i}{2}

Aplicar as propriedades das frações:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 + 3 i}{2} = - (- \frac{1}{2}) - (\frac{3 i}{2})

= - (- \frac{1}{2}) - (\frac{3 i}{2})

Remover os parênteses:

(a) = a, - (- a) = a

= \frac{1}{2} - \frac{3 i}{2}

= \frac{1}{2} - \frac{3}{2} i

Simplificar

3 - 1 \frac{- 1 - 3 i}{2} : \frac{1}{2} + i \frac{3}{2}

3 - 1 \frac{- 1 - 3 i}{2}

3 - 1 = - 1

3 - 1

n - 1 = - 1,

n

é ímpar

3 - 1 = - 1

= - 1

= - 1 \cdot \frac{- 1 - 3 i}{2}

Multiplicar:

1 \cdot \frac{- 1 - 3 i}{2} = \frac{- 1 - 3 i}{2}

= - \frac{- 1 - 3 i}{2}

Reescrever

- \frac{- 1 - 3 i}{2}

na forma complexa padrão:

\frac{1}{2} + \frac{3}{2} i

- \frac{- 1 - 3 i}{2}

Aplicar as propriedades das frações:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 - 3 i}{2} = - (- \frac{1}{2}) - (- \frac{3 i}{2})

= - (- \frac{1}{2}) - (- \frac{3 i}{2})

Aplicar a regra

- (- a) = a

= \frac{1}{2} + \frac{3 i}{2}

= \frac{1}{2} + \frac{3}{2} i

u = - 1, u = \frac{1}{2} - i \frac{3}{2}, u = \frac{1}{2} + i \frac{3}{2}

Substituir na equação

u = sin (x)

sin (x) = - 1, sin (x) = \frac{1}{2} - i \frac{3}{2}, sin (x) = \frac{1}{2} + i \frac{3}{2}

sin (x) = - 1, sin (x) = \frac{1}{2} - i \frac{3}{2}, sin (x) = \frac{1}{2} + i \frac{3}{2}

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

Soluções gerais para

sin (x) = - 1

sin (x)

tabela de periodicidade com ciclo de

2 πn

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = \frac{1}{2} - i \frac{3}{2} :

Sem solução

sin (x) = \frac{1}{2} - i \frac{3}{2}

Sem solu \overset{c}{\c} \tilde{a} o

sin (x) = \frac{1}{2} + i \frac{3}{2} :

Sem solução

sin (x) = \frac{1}{2} + i \frac{3}{2}

Sem solu \overset{c}{\c} \tilde{a} o

Combinar toda as soluções

x = \frac{3 π}{2} + 2 πn

Gráfico

Exemplos populares

solvefor x,sin^2(x)=((1-cos(y)))/2 resolver para

x, sin^{2} (x) = \frac{( 1 - cos ( y ) )}{2}

sin(x)+sin^2(x)+cos(x)+cos^2(x)=0

sin (x) + sin^{2} (x) + cos (x) + cos^{2} (x) = 0

4cos^2(x)-4sin(x)-1=0

4 cos^{2} (x) - 4 sin (x) - 1 = 0

cos^2(x)+1=-2cos(x)

cos^{2} (x) + 1 = - 2 cos (x)

sin^3(x)-2sin(x)=0

sin^{3} (x) - 2 sin (x) = 0