Question

Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°

2 cos²x + 1 = – 3 cos x

Answer 1

2 cos²x + 1 = – 3 cos x

2 cos²x + 3 cos x + 1 = 0

2 cos²x + 2 cos x + cos x + 1 = 0

2 cos x (cos x + 1) + 1(cos x + 1) = 0

(2 cos x + 1)(cos x + 1) = 0

2 cos x + 1 = 0 or cos x + 1 = 0

cos x = `- 1/2` or cos x = – 1

To find the solution of cos x = `- 1/2`

cos x = ` - 1/2`

cos x = `cos (pi - pi/3)`

x = `pi - pi/3`

= `(3pi - pi)/3`

= `(2pi)/3`

General solution is x = `2"n"pi + (2pi)/3`, n ∈ Z

x = `2"n"pi + (2pi)/3`

or

x = `2"n"pi - (2pi)/3`, n ∈ Z

Consider x = `2"n"pi + (2pi)/3`

When n = 0, x = `0 + (2pi)/3 = (2pi)/3` ∈ (0°, 360°)

When n = 1, x = `2pi + (2pi)/3 = (6pi + 2pi)/3 = (8pi)/3` ∉ (0°, 360°)

Consider x = `2"n"pi - (2pi)/3`

When n = 0, x = `0 - (2pi)/3 = - (2pi)/3` ∈ (0°, 360°)

When n = 1, x = `2pi - (2pi)/3 = (6pi - 2pi)/3 = (4pi)/3` ∈ (0°, 360°)

When n = 2, x = `4pi - (2pi)/3 = (12pi - 2pi)/3 = (10pi)/3` ∉ (0°, 360°)

To find the solution of cos x = – 1

cos x = – 1

cos x = cos π

The general solution is

x = 2nπ ± π, n ∈ Z

x = 2nπ + π or x = 2nπ – π, n ∈ Z

Consider x = 2nπ + π

When n = 0 , x = 0 + π = π ∈ (0°, 360°)

When n = 1 , x = 2π + π = 3π ∉ (0°, 360°)

Consider x = 2nπ – π

When n = 0, x = 0 – π ∉ (0°, 360°)

When n = 1, x = 2π – π = π ∈ (0°, 360°)

When n = 2, x = 4π – π = 3π ∉ (0°, 360°)

∴ The required solution are x = `(2pi)/3, (4pi)/3, pi`