sin(2x)=1+cos(2x)

Solución

sin (2 x) = 1 + cos (2 x)

x = \frac{π}{4} + πn, x = πn + \frac{π}{2}

Pasos de solución

sin (2 x) = 1 + cos (2 x)

Restar

1 + cos (2 x)

de ambos lados

sin (2 x) - 1 - cos (2 x) = 0

Re-escribir usando identidades trigonométricas

- 1 + 2 sin (2 x - \frac{π}{4}) = 0

Mover

1

al lado derecho

2 sin (2 x - \frac{π}{4}) = 1

Dividir ambos lados entre

2

sin (2 x - \frac{π}{4}) = \frac{2}{2}

Soluciones generales para

sin (2 x - \frac{π}{4}) = \frac{2}{2}

2 x - \frac{π}{4} = \frac{π}{4} + 2 πn, 2 x - \frac{π}{4} = \frac{3 π}{4} + 2 πn

Resolver

2 x - \frac{π}{4} = \frac{π}{4} + 2 πn : x = \frac{π}{4} + πn

Resolver

2 x - \frac{π}{4} = \frac{3 π}{4} + 2 πn : x = πn + \frac{π}{2}

x = \frac{π}{4} + πn, x = πn + \frac{π}{2}

sin (2 θ) + sin (4 θ) = 0

3 tan (x) - 1 = 0

cos (4 x) = - \frac{1}{2}

16 cos^{2} (x) - 4 = 0

cos (2 x) + cos (x) = - 1