If `sin theta + 2cos theta = -1`, where `0 lt theta lt pi/2`, what is `2 sin theta - cos theta`equal to ?

Options

A. `-1`
B. `1/2`
C. 2
D. 1

Answer

To solve the equation \( \sin \theta + 2 \cos \theta = -1 \) for \( 2 \sin \theta - \cos \theta \), we can follow these steps:

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \sin \theta + 2 \cos \theta = -1 \]

2. **Rearrange the equation:**
\[ \sin \theta = -1 - 2 \cos \theta \]

3. **Square both sides:**
\[ (\sin \theta)^2 = (-1 - 2 \cos \theta)^2 \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 1 - \cos^2 \theta = 1 + 4 \cos^2 \theta + 4 \cos \theta \]

4. **Combine like terms:**
\[ 0 = 5 \cos^2 \theta + 4 \cos \theta \]

5. **Factor the equation:**
\[ \cos \theta (5 \cos \theta + 4) = 0 \]

6. **Solve for \( \cos \theta \):**
- From \( \cos \theta = 0 \), this does not apply since \( 0 < \theta < \frac{\pi}{2} \).
- From \( 5 \cos \theta + 4 = 0 \):
\[ \cos \theta = -\frac{4}{5} \]
This is not valid in the interval \( 0 < \theta < \frac{\pi}{2} \) as cosine is positive in this range.

7. **Revisit the original equation:**
Since the derived conditions do not yield valid solutions in the specified range, we can directly substitute \( \sin \theta \) and \( \cos \theta \) values to find \( 2 \sin \theta - \cos \theta \).

8. **Use the original equation to find \( k = 2 \sin \theta - \cos \theta \):**
- Let \( k = 2 \sin \theta - \cos \theta \).
- Square both sides:
\[ k^2 = (2 \sin \theta - \cos \theta)^2 = 4 \sin^2 \theta - 4 \sin \theta \cos \theta + \cos^2 \theta \]

9. **Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):**
\[ k^2 = 4 \sin^2 \theta + (1 - \sin^2 \theta) - 4 \sin \theta \cos \theta \]
\[ k^2 = 3 \sin^2 \theta + 1 - 4 \sin \theta \cos \theta \]

10. **Combine the equations:**
Add the equations derived from squaring both sides of the original equation and the expression for \( k \):
\[ 1 + k^2 = 1 + 4 \sin^2 \theta + 4 \cos^2 \theta + 4 \sin \theta \cos \theta - 4 \sin \theta \cos \theta \]
Simplifying gives:
\[ k^2 = 4 \]

11. **Solve for \( k \):**
\[ k = \sqrt{4} = 2 \]

### Final Answer:
Thus, \( 2 \sin \theta - \cos \theta = 2 \).