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Popular Trigonometria >

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

Solução

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

Solução

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

Graus

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 38.17270 \dots^{\circ} + 36 0^{\circ} n, x = 141.82729 \dots^{\circ} + 36 0^{\circ} n

Passos da solução

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

Fatorar

sin^{2} (x) cos (x) - sin (x) cos^{3} (x) : sin (x) cos (x) (sin (x) - cos^{2} (x))

sin^{2} (x) cos (x) - sin (x) cos^{3} (x)

Aplicar as propriedades dos expoentes:

a^{b + c} = a^{b} a^{c}

sin (x) cos^{3} (x) = sin (x) cos (x) cos^{2} (x), sin^{2} (x) cos (x) = sin (x) sin (x) cos (x)

= sin (x) sin (x) cos (x) - sin (x) cos (x) cos^{2} (x)

Fatorar o termo comum

sin (x) cos (x)

= sin (x) cos (x) (sin (x) - cos^{2} (x))

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

Resolver cada parte separadamente

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

Soluções gerais para

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Resolver

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

Soluções gerais para

cos (x) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

sin (x) - cos^{2} (x) = 0 : x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

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

Reeecreva usando identidades trigonométricas

- cos^{2} (x) + sin (x)

Utilizar a identidade trigonométrica pitagórica:

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

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

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

- (1 - sin^{2} (x)) : - 1 + sin^{2} (x)

- (1 - sin^{2} (x))

Colocar os parênteses

= - (1) - (- sin^{2} (x))

Aplicar as regras dos sinais

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

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

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

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

Usando o método de substituição

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

Sea:

sin (x) = u

- 1 + u + u^{2} = 0

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

- 1 + u + u^{2} = 0

Escrever na forma padrão

a x^{2} + b x + c = 0

u^{2} + u - 1 = 0

Resolver com a fórmula quadrática

u^{2} + u - 1 = 0

Fórmula geral para equações de segundo grau:

Para

a = 1, b = 1, c = - 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

1^{2} - 4 \cdot 1 \cdot (- 1)

Aplicar a regra

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

Aplicar a regra

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiplicar os números:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Somar:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separe as soluções

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

Multiplicar os números:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

Multiplicar os números:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

As soluções para a equação de segundo grau são:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

Substituir na equação

u = sin (x)

sin (x) = \frac{- 1 + 5}{2}, sin (x) = \frac{- 1 - 5}{2}

sin (x) = \frac{- 1 + 5}{2}, sin (x) = \frac{- 1 - 5}{2}

sin (x) = \frac{- 1 + 5}{2} : x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

sin (x) = \frac{- 1 + 5}{2}

Aplique as propriedades trigonométricas inversas

sin (x) = \frac{- 1 + 5}{2}

Soluções gerais para

sin (x) = \frac{- 1 + 5}{2}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

sin (x) = \frac{- 1 - 5}{2} :

Sem solução

sin (x) = \frac{- 1 - 5}{2}

- 1 \leq sin (x) \leq 1

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

Combinar toda as soluções

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

Combinar toda as soluções

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

Mostrar soluções na forma decimal

x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

Gráfico

Exemplos populares

5sin(θ)-sqrt(3)=3sin(θ)

5 sin (θ) - 3 = 3 sin (θ)

sin^3(x)-4sin(x)=0

sin^{3} (x) - 4 sin (x) = 0

cos(a)=(-82)/(sqrt(140)*\sqrt{138)}

cos (a) = \frac{- 82}{140 \cdot 138}

sec^2(x)-cos(x)=0

sec^{2} (x) - cos (x) = 0

sin(x)*sec(x)=0

sin (x) \cdot sec (x) = 0