Question

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

2 sin²x + 1 = 3 sin x

Answer 1

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`