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solvefor t,y^{11}+9y=18cos(3t)

Solución

resolver para

t, y^{11} + 9 y = 18 cos (3 t)

t = \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}, t = - \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

Pasos de solución

y^{11} + 9 y = 18 cos (3 t)

Intercambiar lados

18 cos (3 t) = y^{11} + 9 y

Dividir ambos lados entre

18

18 cos (3 t) = y^{11} + 9 y

Dividir ambos lados entre

18

\frac{18 cos ( 3 t )}{18} = \frac{y ^{11}}{18} + \frac{9 y}{18}

Simplificar

\frac{18 cos ( 3 t )}{18} = \frac{y ^{11}}{18} + \frac{9 y}{18}

Simplificar

\frac{18 cos ( 3 t )}{18} : cos (3 t)

\frac{18 cos ( 3 t )}{18}

Dividir:

\frac{18}{18} = 1

= cos (3 t)

Simplificar

\frac{y ^{11}}{18} + \frac{9 y}{18} : \frac{y ^{11} + 9 y}{18}

\frac{y ^{11}}{18} + \frac{9 y}{18}

Aplicar la regla

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

= \frac{y ^{11} + 9 y}{18}

cos (3 t) = \frac{y ^{11} + 9 y}{18}

cos (3 t) = \frac{y ^{11} + 9 y}{18}

cos (3 t) = \frac{y ^{11} + 9 y}{18}

Aplicar propiedades trigonométricas inversas

cos (3 t) = \frac{y ^{11} + 9 y}{18}

Soluciones generales para

cos (3 t) = \frac{y ^{11} + 9 y}{18}

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

3 t = arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn, 3 t = - arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn

3 t = arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn, 3 t = - arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn

Resolver

3 t = arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn : t = \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

3 t = arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn

Dividir ambos lados entre

3

3 t = arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn

Dividir ambos lados entre

3

\frac{3 t}{3} = \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

Simplificar

t = \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

t = \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

Resolver

3 t = - arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn : t = - \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

3 t = - arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn

Dividir ambos lados entre

3

3 t = - arccos (\frac{y ^{11} + 9 y}{18}) + 2 πn

Dividir ambos lados entre

3

\frac{3 t}{3} = - \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

Simplificar

t = - \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

t = - \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

t = \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}, t = - \frac{arccos ( \frac{y ^{11} + 9 y}{18} )}{3} + \frac{2 πn}{3}

tan (x) = \frac{0.75}{6}

2 sec (x) - 2 = 0

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

3 cos (x) + 2 sin (x) = 3

sin (x) - 1 = 3 cos (x)