Solve this equation on the interval 0 θ < 2π. Round your answer(s) to two decimal places.
2 sin θ + 3 = 2
(smaller value)
(larger value)
To solve the equation 2 sin θ + 3 = 2 on the interval 0 ≤ θ < 2π, you need to isolate the variable sin θ and then solve for θ.
1. Start by subtracting 3 from both sides of the equation: 2 sin θ + 3 - 3 = 2 - 3.
This simplifies to: 2 sin θ = -1.
2. Divide both sides of the equation by 2 to isolate sin θ: (2 sin θ) / 2 = -1 / 2.
This yields: sin θ = -1/2.
3. Now, you need to determine the angle θ within the given interval that has a sin value of -1/2. One way to find this is by referring to the unit circle or using trigonometric ratios.
Looking at the unit circle, you'll find that sin θ = -1/2 at two angles: π/6 and 5π/6. These are the reference angles in standard position with negative y-coordinate values (sine negative).
Since the interval given is 0 ≤ θ < 2π, only the angles within this interval should be considered.
π/6 is within this interval (as π/6 ≈ 0.524 < 2π), while 5π/6 is not (as 5π/6 ≈ 2.618 > 2π). Therefore, π/6 is the only angle that satisfies the equation within the given interval.
4. Finally, round the solution θ = π/6 to two decimal places: θ ≈ 0.52 (smaller value).
Hence, the solution to the equation 2 sin θ + 3 = 2 on the interval 0 ≤ θ < 2π, rounded to two decimal places, is θ ≈ 0.52.