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Question
Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°
cos 2x = 1 − 3 sin x
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Solution
1 – 2 sin2x = 1 – 3 sinx
2 sin2 x – 3 sin x = 0
sin x(2 sin x – 3) = 0
= sin x = 0 or 2 sin x – 3 = 0
sin x = 0 or sin x = `3/2`
sin x = `3/2` is not possible since sin x ≤ 1
∴ sin x = 0 = sin 0
The general solution is x = nit,
When n = 0, x = 0 ∉ (0°, 360°)
When n = 1, x = π ∈ (0°, 360°)
When n = 2, x = 2π ∉ (0°, 360°)
∴ The required solutions is x = π
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