cos(α_{2})= 1/(sqrt(14)),0<= α_{2}<= 2pi

Solución

cos (α_{2}) = \frac{1}{14}, 0 \leq α_{2} \leq 2 π

α_{2} = 1.30024 \dots, α_{2} = 2 π - 1.30024 \dots

Pasos de solución

cos (α_{2}) = \frac{1}{14}, 0 \leq α_{2} \leq 2 π

Simplificar

\frac{1}{14} : \frac{14}{14}

\frac{1}{14}

Multiplicar por el conjugado

\frac{14}{14}

= \frac{1 \cdot 14}{14 14}

1 \cdot 14 = 14

1414 = 14

1414

Aplicar las leyes de los exponentes:

a a = a

1414 = 14

= 14

= \frac{14}{14}

cos (α_{2}) = \frac{14}{14}

Aplicar propiedades trigonométricas inversas

cos (α_{2}) = \frac{14}{14}

Soluciones generales para

cos (α_{2}) = \frac{14}{14}

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

α_{2} = arccos (\frac{14}{14}) + 2 πn, α_{2} = 2 π - arccos (\frac{14}{14}) + 2 πn

α_{2} = arccos (\frac{14}{14}) + 2 πn, α_{2} = 2 π - arccos (\frac{14}{14}) + 2 πn

Soluciones para el rango

0 \leq α_{2} \leq 2 π

α_{2} = arccos (\frac{14}{14}), α_{2} = 2 π - arccos (\frac{14}{14})

Mostrar soluciones en forma decimal

α_{2} = 1.30024 \dots, α_{2} = 2 π - 1.30024 \dots

sin (2 x + 60) = \frac{1}{2}

tan^{3} (x) - 3 tan (x) = 0

2 cos (x) = 2 cos (2 x)

6 sin (2 x) = 3

- cos (x) = - sin^{2} (x) - 1, 0 \leq x \leq 2 π