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Question
Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°
2 cos2x + 1 = – 3 cos x
Solution
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`
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