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solvefor t,f=cos^2(at)

Solución

resolver para

t, f = cos^{2} (a t)

Solución

t = \frac{arccos ( f )}{a} + \frac{2 πn}{a}, t = - \frac{arccos ( f )}{a} + \frac{2 πn}{a}, t = \frac{arccos ( - f )}{a} + \frac{2 πn}{a}, t = - \frac{arccos ( - f )}{a} + \frac{2 πn}{a}

Pasos de solución

f = cos^{2} (a t)

Intercambiar lados

cos^{2} (a t) = f

Usando el método de sustitución

cos^{2} (a t) = f

Sea:

cos (a t) = u

u^{2} = f

u^{2} = f : u = f, u = - f

u^{2} = f

Para

x^{2} = f (a)

las soluciones son

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

u = f, u = - f

Sustituir en la ecuación

u = cos (a t)

cos (a t) = f, cos (a t) = - f

cos (a t) = f, cos (a t) = - f

cos (a t) = f : t = \frac{arccos ( f )}{a} + \frac{2 πn}{a}, t = - \frac{arccos ( f )}{a} + \frac{2 πn}{a}

cos (a t) = f

Aplicar propiedades trigonométricas inversas

cos (a t) = f

Soluciones generales para

cos (a t) = f

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

a t = arccos (f) + 2 πn, a t = - arccos (f) + 2 πn

a t = arccos (f) + 2 πn, a t = - arccos (f) + 2 πn

Resolver

a t = arccos (f) + 2 πn : t = \frac{arccos ( f )}{a} + \frac{2 πn}{a}; a \neq = 0

a t = arccos (f) + 2 πn

Dividir ambos lados entre

a; a \neq = 0

a t = arccos (f) + 2 πn

Dividir ambos lados entre

a; a \neq = 0

\frac{a t}{a} = \frac{arccos ( f )}{a} + \frac{2 πn}{a}; a \neq = 0

Simplificar

t = \frac{arccos ( f )}{a} + \frac{2 πn}{a}; a \neq = 0

t = \frac{arccos ( f )}{a} + \frac{2 πn}{a}; a \neq = 0

Resolver

a t = - arccos (f) + 2 πn : t = - \frac{arccos ( f )}{a} + \frac{2 πn}{a}; a \neq = 0

a t = - arccos (f) + 2 πn

Dividir ambos lados entre

a; a \neq = 0

a t = - arccos (f) + 2 πn

Dividir ambos lados entre

a; a \neq = 0

\frac{a t}{a} = - \frac{arccos ( f )}{a} + \frac{2 πn}{a}; a \neq = 0

Simplificar

t = - \frac{arccos ( f )}{a} + \frac{2 πn}{a}; a \neq = 0

t = - \frac{arccos ( f )}{a} + \frac{2 πn}{a}; a \neq = 0

t = \frac{arccos ( f )}{a} + \frac{2 πn}{a}, t = - \frac{arccos ( f )}{a} + \frac{2 πn}{a}

cos (a t) = - f : t = \frac{arccos ( - f )}{a} + \frac{2 πn}{a}, t = - \frac{arccos ( - f )}{a} + \frac{2 πn}{a}

cos (a t) = - f

Aplicar propiedades trigonométricas inversas

cos (a t) = - f

Soluciones generales para

cos (a t) = - f

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

a t = arccos (- f) + 2 πn, a t = - arccos (- f) + 2 πn

a t = arccos (- f) + 2 πn, a t = - arccos (- f) + 2 πn

Resolver

a t = arccos (- f) + 2 πn : t = \frac{arccos ( - f )}{a} + \frac{2 πn}{a}; a \neq = 0

a t = arccos (- f) + 2 πn

Dividir ambos lados entre

a; a \neq = 0

a t = arccos (- f) + 2 πn

Dividir ambos lados entre

a; a \neq = 0

\frac{a t}{a} = \frac{arccos ( - f )}{a} + \frac{2 πn}{a}; a \neq = 0

Simplificar

t = \frac{arccos ( - f )}{a} + \frac{2 πn}{a}; a \neq = 0

t = \frac{arccos ( - f )}{a} + \frac{2 πn}{a}; a \neq = 0

Resolver

a t = - arccos (- f) + 2 πn : t = - \frac{arccos ( - f )}{a} + \frac{2 πn}{a}; a \neq = 0

a t = - arccos (- f) + 2 πn

Dividir ambos lados entre

a; a \neq = 0

a t = - arccos (- f) + 2 πn

Dividir ambos lados entre

a; a \neq = 0

\frac{a t}{a} = - \frac{arccos ( - f )}{a} + \frac{2 πn}{a}; a \neq = 0

Simplificar

t = - \frac{arccos ( - f )}{a} + \frac{2 πn}{a}; a \neq = 0

t = - \frac{arccos ( - f )}{a} + \frac{2 πn}{a}; a \neq = 0

t = \frac{arccos ( - f )}{a} + \frac{2 πn}{a}, t = - \frac{arccos ( - f )}{a} + \frac{2 πn}{a}

Combinar toda las soluciones

t = \frac{arccos ( f )}{a} + \frac{2 πn}{a}, t = - \frac{arccos ( f )}{a} + \frac{2 πn}{a}, t = \frac{arccos ( - f )}{a} + \frac{2 πn}{a}, t = - \frac{arccos ( - f )}{a} + \frac{2 πn}{a}

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