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Popular >

(d^2-4)=cos^2(x)

Solución

(d^{2} - 4) = cos^{2} (x)

Solución

x = arccos (d^{2} - 4) + 2 πn, x = - arccos (d^{2} - 4) + 2 πn, x = arccos (- d^{2} - 4) + 2 πn, x = - arccos (- d^{2} - 4) + 2 πn

Pasos de solución

(d^{2} - 4) = cos^{2} (x)

Intercambiar lados

cos^{2} (x) = d^{2} - 4

Usando el método de sustitución

cos^{2} (x) = d^{2} - 4

Sea:

cos (x) = u

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

u^{2} = d^{2} - 4 : u = d^{2} - 4, u = - d^{2} - 4

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

Para

x^{2} = f (a)

las soluciones son

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

u = d^{2} - 4, u = - d^{2} - 4

Sustituir en la ecuación

u = cos (x)

cos (x) = d^{2} - 4, cos (x) = - d^{2} - 4

cos (x) = d^{2} - 4, cos (x) = - d^{2} - 4

cos (x) = d^{2} - 4 : x = arccos (d^{2} - 4) + 2 πn, x = - arccos (d^{2} - 4) + 2 πn

cos (x) = d^{2} - 4

Aplicar propiedades trigonométricas inversas

cos (x) = d^{2} - 4

Soluciones generales para

cos (x) = d^{2} - 4

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

x = arccos (d^{2} - 4) + 2 πn, x = - arccos (d^{2} - 4) + 2 πn

x = arccos (d^{2} - 4) + 2 πn, x = - arccos (d^{2} - 4) + 2 πn

cos (x) = - d^{2} - 4 : x = arccos (- d^{2} - 4) + 2 πn, x = - arccos (- d^{2} - 4) + 2 πn

cos (x) = - d^{2} - 4

Aplicar propiedades trigonométricas inversas

cos (x) = - d^{2} - 4

Soluciones generales para

cos (x) = - d^{2} - 4

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

x = arccos (- d^{2} - 4) + 2 πn, x = - arccos (- d^{2} - 4) + 2 πn

x = arccos (- d^{2} - 4) + 2 πn, x = - arccos (- d^{2} - 4) + 2 πn

Combinar toda las soluciones

x = arccos (d^{2} - 4) + 2 πn, x = - arccos (d^{2} - 4) + 2 πn, x = arccos (- d^{2} - 4) + 2 πn, x = - arccos (- d^{2} - 4) + 2 πn

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sin(x)=(48sin(69))/(47.5)

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2cos^2(3x)+cos(3x)-1=0

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