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Popular Trigonometria >

d=(sin^4(x)-cos^2(x)+5)/(4*cos^2(x))

Solução

d = \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 \cdot cos ^{2} ( x )}

Solução

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

Passos da solução

d = \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )}

Trocar lados

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} = d

Subtrair

d

de ambos os lados

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - d = 0

Simplificar

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - d : \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5 - 4 d cos ^{2} ( x )}{4 cos ^{2} ( x )}

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - d

Converter para fração:

d = \frac{d 4 cos ^{2} ( x )}{4 cos ^{2} ( x )}

= \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - \frac{d \cdot 4 cos ^{2} ( x )}{4 cos ^{2} ( x )}

Já que os denominadores são iguais, combinar as frações:

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

= \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5 - d \cdot 4 cos ^{2} ( x )}{4 cos ^{2} ( x )}

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5 - 4 d cos ^{2} ( x )}{4 cos ^{2} ( x )} = 0

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

sin^{4} (x) - cos^{2} (x) + 5 - 4 d cos^{2} (x) = 0

Reeecreva usando identidades trigonométricas

5 - cos^{2} (x) + sin^{4} (x) - 4 cos^{2} (x) d

Utilizar a identidade trigonométrica pitagórica:

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

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

= 5 - (1 - sin^{2} (x)) + sin^{4} (x) - 4 (1 - sin^{2} (x)) d

Simplificar

5 - (1 - sin^{2} (x)) + sin^{4} (x) - 4 (1 - sin^{2} (x)) d : sin^{4} (x) + sin^{2} (x) + 4 d sin^{2} (x) + 4 - 4 d

5 - (1 - sin^{2} (x)) + sin^{4} (x) - 4 (1 - sin^{2} (x)) d

= 5 - (1 - sin^{2} (x)) + sin^{4} (x) - 4 d (1 - sin^{2} (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)

= 5 - 1 + sin^{2} (x) + sin^{4} (x) - 4 (1 - sin^{2} (x)) d

Expandir

- 4 d (1 - sin^{2} (x)) : - 4 d + 4 d sin^{2} (x)

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

Colocar os parênteses utilizando:

a (b - c) = ab - a c

a = - 4 d, b = 1, c = sin^{2} (x)

= - 4 d \cdot 1 - (- 4 d) sin^{2} (x)

Aplicar as regras dos sinais

- (- a) = a

= - 4 \cdot 1 \cdot d + 4 d sin^{2} (x)

Multiplicar os números:

4 \cdot 1 = 4

= - 4 d + 4 d sin^{2} (x)

= 5 - 1 + sin^{2} (x) + sin^{4} (x) - 4 d + 4 d sin^{2} (x)

Subtrair:

5 - 1 = 4

= sin^{4} (x) + sin^{2} (x) + 4 d sin^{2} (x) + 4 - 4 d

= sin^{4} (x) + sin^{2} (x) + 4 d sin^{2} (x) + 4 - 4 d

4 + sin^{2} (x) + sin^{4} (x) - 4 d + 4 sin^{2} (x) d = 0

Usando o método de substituição

4 + sin^{2} (x) + sin^{4} (x) - 4 d + 4 sin^{2} (x) d = 0

Sea:

sin (x) = u

4 + u^{2} + u^{4} - 4 d + 4 u^{2} d = 0

4 + u^{2} + u^{4} - 4 d + 4 u^{2} d = 0 : u = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

4 + u^{2} + u^{4} - 4 d + 4 u^{2} d = 0

Escrever na forma padrão

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

u^{4} + (1 + 4 d) u^{2} + 4 - 4 d = 0

Reescrever a equação com

v = u^{2}

v^{2} = u^{4}

v^{2} + (1 + 4 d) v + 4 - 4 d = 0

Resolver

v^{2} + (1 + 4 d) v + 4 - 4 d = 0 : v = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, v = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

v^{2} + (1 + 4 d) v + 4 - 4 d = 0

Resolver com a fórmula quadrática

v^{2} + (1 + 4 d) v + 4 - 4 d = 0

Fórmula geral para equações de segundo grau:

Para

a = 1, b = 1 + 4 d, c = 4 - 4 d

v_{1, 2} = \frac{- ( 1 + 4 d ) \pm ( 1 + 4 d ) ^{2} - 4 \cdot 1 \cdot ( 4 - 4 d )}{2 \cdot 1}

v_{1, 2} = \frac{- ( 1 + 4 d ) \pm ( 1 + 4 d ) ^{2} - 4 \cdot 1 \cdot ( 4 - 4 d )}{2 \cdot 1}

Simplificar

(1 + 4 d)^{2} - 4 \cdot 1 \cdot (4 - 4 d) : 16 d^{2} + 24 d - 15

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

Multiplicar os números:

4 \cdot 1 = 4

= (4 d + 1)^{2} - 4 (- 4 d + 4)

Expandir

(1 + 4 d)^{2} - 4 (4 - 4 d) : 16 d^{2} + 24 d - 15

(1 + 4 d)^{2} - 4 (4 - 4 d)

(1 + 4 d)^{2} : 1 + 8 d + 16 d^{2}

Aplique a fórmula do quadrado perfeito:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 1, b = 4 d

= 1^{2} + 2 \cdot 1 \cdot 4 d + (4 d)^{2}

Simplificar

1^{2} + 2 \cdot 1 \cdot 4 d + (4 d)^{2} : 1 + 8 d + 16 d^{2}

1^{2} + 2 \cdot 1 \cdot 4 d + (4 d)^{2}

Aplicar a regra

1^{a} = 1

1^{2} = 1

= 1 + 2 \cdot 1 \cdot 4 d + (4 d)^{2}

2 \cdot 1 \cdot 4 d = 8 d

2 \cdot 1 \cdot 4 d

Multiplicar os números:

2 \cdot 1 \cdot 4 = 8

= 8 d

(4 d)^{2} = 16 d^{2}

(4 d)^{2}

Aplicar as propriedades dos expoentes:

(a \cdot b)^{n} = a^{n} b^{n}

= 4^{2} d^{2}

4^{2} = 16

= 16 d^{2}

= 1 + 8 d + 16 d^{2}

= 1 + 8 d + 16 d^{2}

= 1 + 8 d + 16 d^{2} - 4 (4 - 4 d)

Expandir

- 4 (4 - 4 d) : - 16 + 16 d

- 4 (4 - 4 d)

Colocar os parênteses utilizando:

a (b - c) = ab - a c

a = - 4, b = 4, c = 4 d

= - 4 \cdot 4 - (- 4) \cdot 4 d

Aplicar as regras dos sinais

- (- a) = a

= - 4 \cdot 4 + 4 \cdot 4 d

Multiplicar os números:

4 \cdot 4 = 16

= - 16 + 16 d

= 1 + 8 d + 16 d^{2} - 16 + 16 d

Simplificar

1 + 8 d + 16 d^{2} - 16 + 16 d : 16 d^{2} + 24 d - 15

1 + 8 d + 16 d^{2} - 16 + 16 d

Agrupar termos semelhantes

= 16 d^{2} + 8 d + 16 d + 1 - 16

Somar elementos similares:

8 d + 16 d = 24 d

= 16 d^{2} + 24 d + 1 - 16

Somar/subtrair:

1 - 16 = - 15

= 16 d^{2} + 24 d - 15

= 16 d^{2} + 24 d - 15

= 16 d^{2} + 24 d - 15

v_{1, 2} = \frac{- ( 1 + 4 d ) \pm 16 d ^{2} + 24 d - 15}{2 \cdot 1}

Separe as soluções

v_{1} = \frac{- ( 1 + 4 d ) + 16 d ^{2} + 24 d - 15}{2 \cdot 1}, v_{2} = \frac{- ( 1 + 4 d ) - 16 d ^{2} + 24 d - 15}{2 \cdot 1}

v = \frac{- ( 1 + 4 d ) + 16 d ^{2} + 24 d - 15}{2 \cdot 1} : \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

\frac{- ( 1 + 4 d ) + 16 d ^{2} + 24 d - 15}{2 \cdot 1}

Multiplicar os números:

2 \cdot 1 = 2

= \frac{- ( 4 d + 1 ) + 16 d ^{2} + 24 d - 15}{2}

- (1 + 4 d) : - 1 - 4 d

- (1 + 4 d)

Colocar os parênteses

= - (1) - (4 d)

Aplicar as regras dos sinais

+ (- a) = - a

= - 1 - 4 d

= \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

v = \frac{- ( 1 + 4 d ) - 16 d ^{2} + 24 d - 15}{2 \cdot 1} : \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

\frac{- ( 1 + 4 d ) - 16 d ^{2} + 24 d - 15}{2 \cdot 1}

Multiplicar os números:

2 \cdot 1 = 2

= \frac{- ( 4 d + 1 ) - 16 d ^{2} + 24 d - 15}{2}

- (1 + 4 d) : - 1 - 4 d

- (1 + 4 d)

Colocar os parênteses

= - (1) - (4 d)

Aplicar as regras dos sinais

+ (- a) = - a

= - 1 - 4 d

= \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

As soluções para a equação de segundo grau são:

v = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, v = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

v = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, v = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Substitua

v = u^{2},

solucione para

u

Resolver

u^{2} = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} : u = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

u^{2} = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

Para

x^{2} = f (a)

as soluções são

x = f (a), - f (a)

u = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

Resolver

u^{2} = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} : u = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

u^{2} = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Para

x^{2} = f (a)

as soluções são

x = f (a), - f (a)

u = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

As soluções são

u = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, u = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, u = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Substituir na equação

u = sin (x)

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} : x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

Aplique as propriedades trigonométricas inversas

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

Soluções gerais para

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

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

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} : x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

Aplique as propriedades trigonométricas inversas

sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

Soluções gerais para

sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}

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

x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} : x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Aplique as propriedades trigonométricas inversas

sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Soluções gerais para

sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

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

x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} : x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Aplique as propriedades trigonométricas inversas

sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

Soluções gerais para

sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

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

x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

Combinar toda as soluções

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

Gráfico

Exemplos populares

2+cos^2(x)=5sin(x)

2 + cos^{2} (x) = 5 sin (x)

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

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

cot^5(x)=(-1)/((sqrt(3)))

cot^{5} (x) = \frac{- 1}{( 3 )}

2cos^4(x)cos(x)-cos^5(x)=1

2 cos^{4} (x) cos (x) - cos^{5} (x) = 1

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

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