solvefor x,(d^2-3d+2)y=cos(e^{-x})

Solución

resolver para

x, (d^{2} - 3 d + 2) y = cos (e^{- x})

x = - ln (arccos (y (d^{2} - 3 d + 2)) + 2 πn), x = - ln (- arccos (y (d^{2} - 3 d + 2)) + 2 πn)

Pasos de solución

(d^{2} - 3 d + 2) y = cos (e^{- x})

Intercambiar lados

cos (e^{- x}) = (d^{2} - 3 d + 2) y

Aplicar propiedades trigonométricas inversas

e^{- x} = arccos ((d^{2} - 3 d + 2) y) + 2 πn, e^{- x} = - arccos ((d^{2} - 3 d + 2) y) + 2 πn

Resolver

e^{- x} = arccos ((d^{2} - 3 d + 2) y) + 2 πn : x = - ln (arccos (y (d^{2} - 3 d + 2)) + 2 πn)

Resolver

e^{- x} = - arccos ((d^{2} - 3 d + 2) y) + 2 πn : x = - ln (- arccos (y (d^{2} - 3 d + 2)) + 2 πn)

x = - ln (arccos (y (d^{2} - 3 d + 2)) + 2 πn), x = - ln (- arccos (y (d^{2} - 3 d + 2)) + 2 πn)

tan (θ) = \frac{6}{12}

(tan (x) + 1)^{2} + (1 + \frac{1}{tan ( x )})^{2} = 100

8 cos (x + 1 0^{\circ}) = 5

cos^{2} (x) + cos (- x) = 0

x, ln (y) + y^{2} = - cos (x)