# solve trigonometry equation

2cos2θ + 3cosθ + 1 = 0

Double Angle Formulas cos2θ = 2cos2θ - 1

2[2cos2θ - 1] + 3cosθ + 1 = 0

4cos2θ - 2 + 3cosθ + 1 = 0

4cos2θ + 3cosθ - 1 = 0

Let cosθ = x

4x2 + 3x - 1 = 0

Now solve the equation using the factor method.

4x2 + 4x - x - 1 = 0

4x(x + 1) -1(x + 1) = 0

(4x - 1)(x + 1) = 0

4x - 1 = 0 or x + 1 = 0.

4x - 1 = 0 then x = 1/4

x + 1 = 0 then x = - 1

But x = cosθ

If x = 1/4 then cosθ = 1/4 ⇒ θ = arc cos(1/4) ⇒ θ = 75°

If x = -1 then cosθ = -1 ⇒ θ = arc cos(-1) ⇒ θ = 180°

Therefore θ = 75° or 180°.

cos θ = - 1 or cos θ = 1/4.

• cos θ = - 1.

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

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

cos θ = cos π.

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

θ = 2nπ ± π, where n is an integer.

• cos θ = 1/4.

⇒ θ = cos-1 (1/4).

cos θ = cos (cos-1 (1/4))

The genaral solution of cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

θ = 2nπ ± cos-1 (1/4), where n is an integer.

The solutions are θ = 2nπ ± π and θ = 2nπ ± cos-1 (1/4), where n is an integer.