solve 3cos(2x) + cos(x) = 2

Question

use double angle trig identities.

Answer 1

The trig equation is 3Cos(2x ) + Cosx  = 2.

We know that the double angle trig identity

Apply zero product property.

Comments

cos(x) = - 1 or cos(x) = 5/6.

x = cos^{- 1}(- 1) or x = cos^{- 1}(5/6).

The function cos (x) has a period of 2π, first find all solutions in the interval (0, 2π).

The function cos (x) is negative in second and third quadrant.

• cos(x) = - 1.

In second Quadrant, π/2 ≤ x ≤ π.

- 1 = - cos (0) = cos (π - 0) = cos (π).

In third Quadrant, π ≤ x ≤ 3π/2.

- 1 = - cos (0) = cos (π + 0) = cos (π).

So, the general solutions is x = 2n₁π + π, where n₁ ∈ Z.

• cos(x) = 5/6.

x = cos^{- 1}(5/6).

So, the general solution is x = 2n₂π ± cos^{- 1}(5/6), where n₂ ∈ Z.

The general solutions of are x = 2n₁π + π and x = 2n₂π ± cos^{- 1}(5/6), where n₁, n₂∈ Z.