solvefor t,x=cos(t)+cos(2t)

Solución

resolver para

t, x = cos (t) + cos (2 t)

t = arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn

Pasos de solución

x = cos (t) + cos (2 t)

Intercambiar lados

cos (t) + cos (2 t) = x

Restar

x

de ambos lados

cos (t) + cos (2 t) - x = 0

Re-escribir usando identidades trigonométricas

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

Usando el método de sustitución

cos (t) = \frac{- 1 + 8 x + 9}{4}, cos (t) = \frac{- 1 - 8 x + 9}{4}

cos (t) = \frac{- 1 + 8 x + 9}{4} : t = arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn

cos (t) = \frac{- 1 - 8 x + 9}{4} : t = arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn

Combinar toda las soluciones

t = arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 + 8 x + 9}{4}) + 2 πn, t = arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn, t = - arccos (\frac{- 1 - 8 x + 9}{4}) + 2 πn

x, sin (x) = - 1

9 - 9 sin (θ) = 6 cos^{2} (θ)

cot (2 θ) = 0

sec (2 x) = - 2

cos (x) = \frac{5}{7}