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(cos^3(x))/(sin(x))=cot(x)

Solución

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

Solución

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

Grados

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Pasos de solución

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

Restar

cot (x)

de ambos lados

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

Simplificar

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

\frac{cos ^{3} ( x )}{sin ( x )} - cot (x)

Convertir a fracción:

cot (x) = \frac{cot ( x ) sin ( x )}{sin ( x )}

= \frac{cos ^{3} ( x )}{sin ( x )} - \frac{cot ( x ) sin ( x )}{sin ( x )}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{cos ^{3} ( x ) - cot ( x ) sin ( x )}{sin ( x )}

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

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

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

Re-escribir usando identidades trigonométricas

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

Utilizar la identidad trigonométrica básica:

cot (x) = \frac{cos ( x )}{sin ( x )}

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

\frac{cos ( x )}{sin ( x )} sin (x) = cos (x)

\frac{cos ( x )}{sin ( x )} sin (x)

Multiplicar fracciones:

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

= \frac{cos ( x ) sin ( x )}{sin ( x )}

Eliminar los terminos comunes:

sin (x)

= cos (x)

= cos^{3} (x) - cos (x)

- cos (x) + cos^{3} (x) = 0

Usando el método de sustitución

- cos (x) + cos^{3} (x) = 0

Sea:

cos (x) = u

- u + u^{3} = 0

- u + u^{3} = 0 : u = 0, u = - 1, u = 1

- u + u^{3} = 0

Factorizar

- u + u^{3} : u (u + 1) (u - 1)

- u + u^{3}

Factorizar el termino común

u : u (u^{2} - 1)

u^{3} - u

Aplicar las leyes de los exponentes:

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

u^{3} = u^{2} u

= u^{2} u - u

Factorizar el termino común

u

= u (u^{2} - 1)

= u (u^{2} - 1)

Factorizar

u^{2} - 1 : (u + 1) (u - 1)

u^{2} - 1

Reescribir

1

como

1^{2}

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

Aplicar la siguiente regla para binomios al cuadrado:

x^{2} - y^{2} = (x + y) (x - y)

u^{2} - 1^{2} = (u + 1) (u - 1)

= (u + 1) (u - 1)

= u (u + 1) (u - 1)

u (u + 1) (u - 1) = 0

Utilizando el teorema de factor cero: si

ab = 0

entonces

a = 0

b = 0

u = 0 or u + 1 = 0 or u - 1 = 0

Resolver

u + 1 = 0 : u = - 1

u + 1 = 0

Mover

1

al lado derecho

u + 1 = 0

Restar

1

de ambos lados

u + 1 - 1 = 0 - 1

Simplificar

u = - 1

u = - 1

Resolver

u - 1 = 0 : u = 1

u - 1 = 0

Mover

1

al lado derecho

u - 1 = 0

Sumar

1

a ambos lados

u - 1 + 1 = 0 + 1

Simplificar

u = 1

u = 1

Las soluciones son

u = 0, u = - 1, u = 1

Sustituir en la ecuación

u = cos (x)

cos (x) = 0, cos (x) = - 1, cos (x) = 1

cos (x) = 0, cos (x) = - 1, cos (x) = 1

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

cos (x) = 0

Soluciones generales para

cos (x) = 0

cos (x)

tabla de valores periódicos con

2 πn

intervalos:

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

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

Soluciones generales para

cos (x) = - 1

cos (x)

tabla de valores periódicos con

2 πn

intervalos:

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 = π + 2 πn

x = π + 2 πn

cos (x) = 1 : x = 2 πn

cos (x) = 1

Soluciones generales para

cos (x) = 1

cos (x)

tabla de valores periódicos con

2 πn

intervalos:

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 = 0 + 2 πn

x = 0 + 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

Combinar toda las soluciones

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

Siendo que la ecuación esta indefinida para:

π + 2 πn, 2 πn

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

Ejemplos populares

tan(2x)= 3/4

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sin(x)=1+cos(x)

sin (x) = 1 + cos (x)

sqrt(2)sin(x)=sqrt(1+(cos(2x))/(sin(x)))

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

sin(x)=(sqrt(21))/5

sin (x) = \frac{21}{5}

tan(θ/2-pi/6)=1

tan (\frac{θ}{2} - \frac{π}{6}) = 1