¿Cómo resuelvo está ecuación trigonométrica? Cos(3x)+sen x= cos x Gracias.?

Do you know this identity?

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) → suppose that: a = b = x

cos(x + x) = cos(x).cos(x) - sin(x).sin(x)

cos(2x) = cos²(x) - sin²(x) ← memorize this result as (1)

Do you know this identity?

sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → suppose that: a = b = x

sin(x + x) = sin(x).cos(x) + cos(x).sin(x)

sin(2x) = 2.sin(x).cos(x) ← memorize this result as (2)

You know this identity?

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) → suppose that: a = 2x and that: b = x

cos(2x + x) = cos(2x).cos(x) - sin(2x).sin(x)

cos(3x) = cos(2x).cos(x) - sin(2x).sin(x) ← memorize this result as (3)

Your equation:

cos(3x) + sin(x) = cos(x) → recall (3)

cos(2x).cos(x) - sin(2x).sin(x) + sin(x) = cos(x)

cos(2x).cos(x) - sin(2x).sin(x) + sin(x) - cos(x) = 0

cos(x).[cos(2x) - 1] + sin(x).[1 - sin(2x)] = 0 → recall (1)

cos(x).[cos²(x) - sin²(x) - 1] + sin(x).[1 - sin(2x)] = 0 → recall (2)

cos(x).[cos²(x) - sin²(x) - 1] + sin(x).[1 - 2.sin(x).cos(x)] = 0

cos(x).[cos²(x) - sin²(x) - {cos²(x) + sin²(x)}] + sin(x).[1 - 2.sin(x).cos(x)] = 0

cos(x).[cos²(x) - sin²(x) - cos²(x) - sin²(x)] + sin(x).[1 - 2.sin(x).cos(x)] = 0

cos(x).[- 2.sin²(x)] + sin(x).[1 - 2.sin(x).cos(x)] = 0

- 2.cos(x).sin²(x) + sin(x) - 2.cos(x).sin²(x) = 0

- 4.cos(x).sin²(x) + sin(x) = 0

4.cos(x).sin²(x) - sin(x) = 0

sin(x).[4.cos(x).sin(x) - 1] = 0

First case:

sin(x) = 0

x = kπ → where k is an integer

Second case:

4.cos(x).sin(x) - 1 = 0

4.cos(x).sin(x) = 1

2.cos(x).sin(x) = 1/2 → recall (2)

sin(2x) = 1/2 ← the corresponding angle is (π/6) or [π - (π/6)]

First possibility:

2x = (π/6) + 2kπ → where k is an integer

x = (π/12) + kπ

Second possibility:

2x = [π - (π/6)] + 2kπ → where k is an integer

2x = (5π/6) + 2kπ

x = (5π/12) + kπ