If `sintheta=K, -1<=K<=1` then number of values of theta for same value of K in`[0,2pi]`

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the number of values of θ in the interval [0, 2π] for the equation sin(θ) = K, where K is a constant such that -1 ≤ K ≤ 1, we can follow these steps:

### Step 1: Understand the range of sine function
The sine function, sin(θ), oscillates between -1 and 1. Therefore, for any value of K within the range of -1 to 1, there will be corresponding angles θ that satisfy the equation sin(θ) = K.

**Hint:** Remember that the sine function is periodic and has a specific range.

### Step 2: Identify the quadrants
The sine function is positive in the first and second quadrants (0 to π) and negative in the third and fourth quadrants (π to 2π).

**Hint:** Think about where the sine function is positive and negative on the unit circle.

### Step 3: Determine the number of solutions in [0, 2π]
1. For K = 0: sin(θ) = 0 has solutions at θ = 0, π, and 2π. However, since we are looking for values in the interval [0, 2π], we consider θ = 0 and θ = π (2π is the same as 0).

2. For 0 < K < 1: There will be two solutions in the interval [0, 2π]. One solution will be in the first quadrant and the other in the second quadrant.

3. For -1 < K < 0: Similarly, there will be two solutions in the interval [0, 2π]. One solution will be in the third quadrant and the other in the fourth quadrant.

4. For K = -1: sin(θ) = -1 has a single solution at θ = 3π/2.

**Hint:** Consider how many angles correspond to the same sine value based on the unit circle.

### Conclusion:
- For K = 0: 2 solutions (0 and π)
- For K = 1: 1 solution (π/2)
- For K = -1: 1 solution (3π/2)
- For 0 < K < 1: 2 solutions
- For -1 < K < 0: 2 solutions

Thus, for any K in the range -1 < K < 1, there are 2 values of θ in the interval [0, 2π] that satisfy sin(θ) = K.

### Final Answer:
The number of values of θ for the same value of K in the interval [0, 2π] is 2 (for K in the range -1 < K < 1). For K = 0, there are 2 solutions, and for K = ±1, there is 1 solution each.

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