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Question
Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°
2 sin2x + 1 = 3 sin x
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Solution
2 sin2x – 3 sin x + 1 = 0
2 sin2x – 2 sin x – sin x + 1 = 0
2 sin x (sin x – 1) – 1(sin x – 1) = 0
(2 sin x – 1)(sin x – 1) = 0
2 sin x – 1 = 0 or sin x – 1 = 0
sin x = `1/2` or sin x = 1
To find the solution of sin x = `1/2`
sin x = `1/2`
sin x = `sin (pi/6)`
The general solution is x = `"n"pi + (-1)^"n" pi/6`, n ∈ z
When n = 0, x = `0 + pi/6 = pi/6` ∈ (0°, 360°)
When n = 1, x = `pi - pi/6 = (6pi - pi)/6 = (5pi)/6` ∈ (0°, 360°)
When n = 2, x = `2pi + pi/6 = (12pi - pi)/6 = (13pi)/6` ∉ (0°, 360°)
To find the solution od sin x = 1
sin x = 1
sin x = `sin (pi/2)`
The general solution is x = `"n"pi + (-1)^"n" pi/2`, n ∈ z
When n = 0, x = `0 + pi/2 = pi/2` ∈ (0°, 360°)
When n = 1, x = `pi - pi/2 = (2pi - pi)/2 = pi/2` ∈ (0°, 360°)
When n = 2, x = `2pi + pi/2 = (4pi - pi)/2 = (5pi)/2` ∉ (0°, 360°)
∴ The required solutions are x = `pi/6, (5pi)/6, pi/2`
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