Find all solutions in the interval [0, 2π).
cos2x + 2 cos x + 1 = 0 (3 points)

x = 2π
x = π
x = pi divided by four. , seven pi divided by four.
x = pi divided by two. , three pi divided by two.

First you must know that for remarkable angles: cos (0) = 1, cos (π) = - 1, cos (π / 2) = 0, cos (3π / 2) = 0, cos (2π) = 1. Then, by simple substitution in the given formula, you can find the solutions of x. Which for the interval [0, 2π) are: x = π, x = pi divided by two and x = three pi divided by two.Attached solution.

Answer Link

Find all solutions in the interval [0, 2π). cos2x + 2 cos x + 1 = 0 (3 points) x = 2π x = π x = pi divided by four. , seven pi divided by four. x = pi divided by two. , three pi divided by two.

Respuesta :

Otras preguntas

Find all solutions in the interval [0, 2π).
cos2x + 2 cos x + 1 = 0 (3 points)

x = 2π
x = π
x = pi divided by four. , seven pi divided by four.
x = pi divided by two. , three pi divided by two.