English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometria >

sin^2(x)cos(x)= 2/(3pi)

Solução

sin^{2} (x) cos (x) = \frac{2}{3 π}

Solução

x = 1.34554 \dots + 2 πn, x = 2 π - 1.34554 \dots + 2 πn, x = 0.51672 \dots + 2 πn, x = 2 π - 0.51672 \dots + 2 πn

Graus

x = 77.09423 \dots^{\circ} + 36 0^{\circ} n, x = 282.90576 \dots^{\circ} + 36 0^{\circ} n, x = 29.60626 \dots^{\circ} + 36 0^{\circ} n, x = 330.39373 \dots^{\circ} + 36 0^{\circ} n

Passos da solução

sin^{2} (x) cos (x) = \frac{2}{3 π}

Subtrair

\frac{2}{3 π}

de ambos os lados

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

Simplificar

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

sin^{2} (x) cos (x) - \frac{2}{3 π}

Converter para fração:

sin^{2} (x) cos (x) = \frac{sin ^{2} ( x ) cos ( x ) 3 π}{3 π}

= \frac{sin ^{2} ( x ) cos ( x ) \cdot 3 π}{3 π} - \frac{2}{3 π}

Já que os denominadores são iguais, combinar as frações:

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

= \frac{sin ^{2} ( x ) cos ( x ) \cdot 3 π - 2}{3 π}

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

3 π sin^{2} (x) cos (x) - 2 = 0

Reeecreva usando identidades trigonométricas

- 2 + 3 cos (x) sin^{2} (x) π

Utilizar a identidade trigonométrica pitagórica:

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

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

= - 2 + 3 cos (x) (1 - cos^{2} (x)) π

- 2 + (1 - cos^{2} (x)) \cdot 3 cos (x) π = 0

Usando o método de substituição

- 2 + (1 - cos^{2} (x)) \cdot 3 cos (x) π = 0

Sea:

cos (x) = u

- 2 + (1 - u^{2}) \cdot 3 u π = 0

- 2 + (1 - u^{2}) \cdot 3 u π = 0 : u \approx 0.22334 \dots, u \approx 0.86944 \dots, u \approx - 1.09278 \dots

- 2 + (1 - u^{2}) \cdot 3 u π = 0

Expandir

- 2 + (1 - u^{2}) \cdot 3 u π : - 2 + 3 π u - 3 π u^{3}

- 2 + (1 - u^{2}) \cdot 3 u π

= - 2 + 3 π u (1 - u^{2})

Expandir

3 u π (1 - u^{2}) : 3 π u - 3 π u^{3}

3 u π (1 - u^{2})

Colocar os parênteses utilizando:

a (b - c) = ab - a c

a = 3 u π, b = 1, c = u^{2}

= 3 u π 1 - 3 u π u^{2}

= 3 \cdot 1 π u - 3 π u^{2} u

Simplificar

3 \cdot 1 π u - 3 π u^{2} u : 3 π u - 3 π u^{3}

3 \cdot 1 π u - 3 π u^{2} u

3 \cdot 1 π u = 3 π u

3 \cdot 1 π u

Multiplicar os números:

3 \cdot 1 = 3

= 3 π u

3 π u^{2} u = 3 π u^{3}

3 π u^{2} u

Aplicar as propriedades dos expoentes:

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

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

= 3 π u^{2 + 1}

Somar:

2 + 1 = 3

= 3 π u^{3}

= 3 π u - 3 π u^{3}

= 3 π u - 3 π u^{3}

= - 2 + 3 π u - 3 π u^{3}

- 2 + 3 π u - 3 π u^{3} = 0

Escrever na forma padrão

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- 3 π u^{3} + 3 π u - 2 = 0

Encontrar uma solução para

- 9.42477 \dots u^{3} + 9.42477 \dots u - 2 = 0

utilizando o método de Newton-Raphson:

u \approx 0.22334 \dots

- 9.42477 \dots u^{3} + 9.42477 \dots u - 2 = 0

Definição de método de Newton-Raphson

f (u) = - 9.42477 \dots u^{3} + 9.42477 \dots u - 2

Encontrar

f^{^{'}} (u) : - 28.27433 \dots u^{2} + 9.42477 \dots

\frac{d}{d u} (- 9.42477 \dots u^{3} + 9.42477 \dots u - 2)

Aplicar a regra da soma/diferença:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (9.42477 \dots u^{3}) + \frac{d}{d u} (9.42477 \dots u) - \frac{d}{d u} (2)

\frac{d}{d u} (9.42477 \dots u^{3}) = 28.27433 \dots u^{2}

\frac{d}{d u} (9.42477 \dots u^{3})

Retirar a constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 9.42477 \dots \frac{d}{d u} (u^{3})

Aplicar a regra da potência:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 9.42477 \dots \cdot 3 u^{3 - 1}

Simplificar

= 28.27433 \dots u^{2}

\frac{d}{d u} (9.42477 \dots u) = 9.42477 \dots

\frac{d}{d u} (9.42477 \dots u)

Retirar a constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 9.42477 \dots \frac{d u}{d u}

Aplicar a regra da derivação:

\frac{d u}{d u} = 1

= 9.42477 \dots \cdot 1

Simplificar

= 9.42477 \dots

\frac{d}{d u} (2) = 0

\frac{d}{d u} (2)

Derivada de uma constante:

\frac{d}{d x} (a) = 0

= 0

= - 28.27433 \dots u^{2} + 9.42477 \dots - 0

Simplificar

= - 28.27433 \dots u^{2} + 9.42477 \dots

Seja

u_{0} = 0

Calcular

u_{n + 1}

até que

Δ u_{n + 1} < 0.000001

u_{1} = 0.21220 \dots : Δ u_{1} = 0.21220 \dots

f (u_{0}) = - 9.42477 \dots \cdot 0^{3} + 9.42477 \dots \cdot 0 - 2 = - 2

f^{^{'}} (u_{0}) = - 28.27433 \dots \cdot 0^{2} + 9.42477 \dots = 9.42477 \dots

u_{1} = 0.21220 \dots

Δ u_{1} = ∣ 0.21220 \dots - 0 ∣ = 0.21220 \dots

Δ u_{1} = 0.21220 \dots

u_{2} = 0.22325 \dots : Δ u_{2} = 0.01104 \dots

f (u_{1}) = - 9.42477 \dots \cdot 0.21220 \dots^{3} + 9.42477 \dots \cdot 0.21220 \dots - 2 = - 0.09006 \dots

f^{^{'}} (u_{1}) = - 28.27433 \dots \cdot 0.21220 \dots^{2} + 9.42477 \dots = 8.15153 \dots

u_{2} = 0.22325 \dots

Δ u_{2} = ∣ 0.22325 \dots - 0.21220 \dots ∣ = 0.01104 \dots

Δ u_{2} = 0.01104 \dots

u_{3} = 0.22334 \dots : Δ u_{3} = 0.00009 \dots

f (u_{2}) = - 9.42477 \dots \cdot 0.22325 \dots^{3} + 9.42477 \dots \cdot 0.22325 \dots - 2 = - 0.00074 \dots

f^{^{'}} (u_{2}) = - 28.27433 \dots \cdot 0.22325 \dots^{2} + 9.42477 \dots = 8.01550 \dots

u_{3} = 0.22334 \dots

Δ u_{3} = ∣ 0.22334 \dots - 0.22325 \dots ∣ = 0.00009 \dots

Δ u_{3} = 0.00009 \dots

u_{4} = 0.22334 \dots : Δ u_{4} = 6.80778 E - 9

f (u_{3}) = - 9.42477 \dots \cdot 0.22334 \dots^{3} + 9.42477 \dots \cdot 0.22334 \dots - 2 = - 5.45598 E - 8

f^{^{'}} (u_{3}) = - 28.27433 \dots \cdot 0.22334 \dots^{2} + 9.42477 \dots = 8.01432 \dots

u_{4} = 0.22334 \dots

Δ u_{4} = ∣ 0.22334 \dots - 0.22334 \dots ∣ = 6.80778 E - 9

Δ u_{4} = 6.80778 E - 9

u \approx 0.22334 \dots

Aplicar a divisão longa

Eq u a t i o n 0 : \frac{- 3 π u ^{3} + 3 π u - 2}{u - 0.22334 \dots} = - 9.42477 \dots u^{2} - 2.10500 \dots u + 8.95462 \dots

- 9.42477 \dots u^{2} - 2.10500 \dots u + 8.95462 \dots \approx 0

Encontrar uma solução para

- 9.42477 \dots u^{2} - 2.10500 \dots u + 8.95462 \dots = 0

utilizando o método de Newton-Raphson:

u \approx 0.86944 \dots

- 9.42477 \dots u^{2} - 2.10500 \dots u + 8.95462 \dots = 0

Definição de método de Newton-Raphson

f (u) = - 9.42477 \dots u^{2} - 2.10500 \dots u + 8.95462 \dots

Encontrar

f^{^{'}} (u) : - 18.84955 \dots u - 2.10500 \dots

\frac{d}{d u} (- 9.42477 \dots u^{2} - 2.10500 \dots u + 8.95462 \dots)

Aplicar a regra da soma/diferença:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (9.42477 \dots u^{2}) - \frac{d}{d u} (2.10500 \dots u) + \frac{d}{d u} (8.95462 \dots)

\frac{d}{d u} (9.42477 \dots u^{2}) = 18.84955 \dots u

\frac{d}{d u} (9.42477 \dots u^{2})

Retirar a constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 9.42477 \dots \frac{d}{d u} (u^{2})

Aplicar a regra da potência:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 9.42477 \dots \cdot 2 u^{2 - 1}

Simplificar

= 18.84955 \dots u

\frac{d}{d u} (2.10500 \dots u) = 2.10500 \dots

\frac{d}{d u} (2.10500 \dots u)

Retirar a constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 2.10500 \dots \frac{d u}{d u}

Aplicar a regra da derivação:

\frac{d u}{d u} = 1

= 2.10500 \dots \cdot 1

Simplificar

= 2.10500 \dots

\frac{d}{d u} (8.95462 \dots) = 0

\frac{d}{d u} (8.95462 \dots)

Derivada de uma constante:

\frac{d}{d x} (a) = 0

= 0

= - 18.84955 \dots u - 2.10500 \dots + 0

Simplificar

= - 18.84955 \dots u - 2.10500 \dots

Seja

u_{0} = 4

Calcular

u_{n + 1}

até que

Δ u_{n + 1} < 0.000001

u_{1} = 2.06121 \dots : Δ u_{1} = 1.93878 \dots

f (u_{0}) = - 9.42477 \dots \cdot 4^{2} - 2.10500 \dots \cdot 4 + 8.95462 \dots = - 150.26184 \dots

f^{^{'}} (u_{0}) = - 18.84955 \dots \cdot 4 - 2.10500 \dots = - 77.50323 \dots

u_{1} = 2.06121 \dots

Δ u_{1} = ∣ 2.06121 \dots - 4 ∣ = 1.93878 \dots

Δ u_{1} = 1.93878 \dots

u_{2} = 1.19627 \dots : Δ u_{2} = 0.86494 \dots

f (u_{1}) = - 9.42477 \dots \cdot 2.06121 \dots^{2} - 2.10500 \dots \cdot 2.06121 \dots + 8.95462 \dots = - 35.42655 \dots

f^{^{'}} (u_{1}) = - 18.84955 \dots \cdot 2.06121 \dots - 2.10500 \dots = - 40.95805 \dots

u_{2} = 1.19627 \dots

Δ u_{2} = ∣ 1.19627 \dots - 2.06121 \dots ∣ = 0.86494 \dots

Δ u_{2} = 0.86494 \dots

u_{3} = 0.91027 \dots : Δ u_{3} = 0.28599 \dots

f (u_{2}) = - 9.42477 \dots \cdot 1.19627 \dots^{2} - 2.10500 \dots \cdot 1.19627 \dots + 8.95462 \dots = - 7.05099 \dots

f^{^{'}} (u_{2}) = - 18.84955 \dots \cdot 1.19627 \dots - 2.10500 \dots = - 24.65418 \dots

u_{3} = 0.91027 \dots

Δ u_{3} = ∣ 0.91027 \dots - 1.19627 \dots ∣ = 0.28599 \dots

Δ u_{3} = 0.28599 \dots

u_{4} = 0.87025 \dots : Δ u_{4} = 0.04001 \dots

f (u_{3}) = - 9.42477 \dots \cdot 0.91027 \dots^{2} - 2.10500 \dots \cdot 0.91027 \dots + 8.95462 \dots = - 0.77088 \dots

f^{^{'}} (u_{3}) = - 18.84955 \dots \cdot 0.91027 \dots - 2.10500 \dots = - 19.26328 \dots

u_{4} = 0.87025 \dots

Δ u_{4} = ∣ 0.87025 \dots - 0.91027 \dots ∣ = 0.04001 \dots

Δ u_{4} = 0.04001 \dots

u_{5} = 0.86944 \dots : Δ u_{5} = 0.00081 \dots

f (u_{4}) = - 9.42477 \dots \cdot 0.87025 \dots^{2} - 2.10500 \dots \cdot 0.87025 \dots + 8.95462 \dots = - 0.01509 \dots

f^{^{'}} (u_{4}) = - 18.84955 \dots \cdot 0.87025 \dots - 2.10500 \dots = - 18.50896 \dots

u_{5} = 0.86944 \dots

Δ u_{5} = ∣ 0.86944 \dots - 0.87025 \dots ∣ = 0.00081 \dots

Δ u_{5} = 0.00081 \dots

u_{6} = 0.86944 \dots : Δ u_{6} = 3.38898 E - 7

f (u_{5}) = - 9.42477 \dots \cdot 0.86944 \dots^{2} - 2.10500 \dots \cdot 0.86944 \dots + 8.95462 \dots = - 6.26743 E - 6

f^{^{'}} (u_{5}) = - 18.84955 \dots \cdot 0.86944 \dots - 2.10500 \dots = - 18.49358 \dots

u_{6} = 0.86944 \dots

Δ u_{6} = ∣ 0.86944 \dots - 0.86944 \dots ∣ = 3.38898 E - 7

Δ u_{6} = 3.38898 E - 7

u \approx 0.86944 \dots

Aplicar a divisão longa

Eq u a t i o n 0 : \frac{- 9.42477 \dots u ^{2} - 2.10500 \dots u + 8.95462 \dots}{u - 0.86944 \dots} = - 9.42477 \dots u - 10.29929 \dots

- 9.42477 \dots u - 10.29929 \dots \approx 0

u \approx - 1.09278 \dots

As soluções são

u \approx 0.22334 \dots, u \approx 0.86944 \dots, u \approx - 1.09278 \dots

Substituir na equação

u = cos (x)

cos (x) \approx 0.22334 \dots, cos (x) \approx 0.86944 \dots, cos (x) \approx - 1.09278 \dots

cos (x) \approx 0.22334 \dots, cos (x) \approx 0.86944 \dots, cos (x) \approx - 1.09278 \dots

cos (x) = 0.22334 \dots : x = arccos (0.22334 \dots) + 2 πn, x = 2 π - arccos (0.22334 \dots) + 2 πn

cos (x) = 0.22334 \dots

Aplique as propriedades trigonométricas inversas

cos (x) = 0.22334 \dots

Soluções gerais para

cos (x) = 0.22334 \dots

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (0.22334 \dots) + 2 πn, x = 2 π - arccos (0.22334 \dots) + 2 πn

x = arccos (0.22334 \dots) + 2 πn, x = 2 π - arccos (0.22334 \dots) + 2 πn

cos (x) = 0.86944 \dots : x = arccos (0.86944 \dots) + 2 πn, x = 2 π - arccos (0.86944 \dots) + 2 πn

cos (x) = 0.86944 \dots

Aplique as propriedades trigonométricas inversas

cos (x) = 0.86944 \dots

Soluções gerais para

cos (x) = 0.86944 \dots

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (0.86944 \dots) + 2 πn, x = 2 π - arccos (0.86944 \dots) + 2 πn

x = arccos (0.86944 \dots) + 2 πn, x = 2 π - arccos (0.86944 \dots) + 2 πn

cos (x) = - 1.09278 \dots :

Sem solução

cos (x) = - 1.09278 \dots

- 1 \leq cos (x) \leq 1

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

Combinar toda as soluções

x = arccos (0.22334 \dots) + 2 πn, x = 2 π - arccos (0.22334 \dots) + 2 πn, x = arccos (0.86944 \dots) + 2 πn, x = 2 π - arccos (0.86944 \dots) + 2 πn

Mostrar soluções na forma decimal

x = 1.34554 \dots + 2 πn, x = 2 π - 1.34554 \dots + 2 πn, x = 0.51672 \dots + 2 πn, x = 2 π - 0.51672 \dots + 2 πn

Gráfico

Exemplos populares

tan(φ)=-1/(sqrt(6))

tan (φ) = - \frac{1}{6}

sin(θ)= 4/5 cos(θ)

sin (θ) = \frac{4}{5} cos (θ)

19= 1/2*7.9*6.2sin(x)

19 = \frac{1}{2} \cdot 7.9 \cdot 6.2 sin (x)

2tan(60-x)=tan(x)

2 tan (60 - x) = tan (x)

cos(x)=-0.71

cos (x) = - 0.71