If `sin theta + 2cos theta` = 1, then what is `2sin theta - cos theta` equal to?

Options

A. 0
B. 1
C. 2
D. 4

Answer

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

### Step 1: Start with the given equation
We have:
\[ \sin \theta + 2 \cos \theta = 1 \]

### Step 2: Square both sides
Squaring both sides gives:
\[ (\sin \theta + 2 \cos \theta)^2 = 1^2 \]
Expanding the left side using the identity \( (A + B)^2 = A^2 + B^2 + 2AB \):
\[ \sin^2 \theta + 4 \cos^2 \theta + 4 \sin \theta \cos \theta = 1 \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ \sin^2 \theta + 4 \cos^2 \theta + 4 \sin \theta \cos \theta - 1 = 0 \]

### Step 4: Define a new variable
Let \( x = 2 \sin \theta - \cos \theta \). We will square this expression as well:
\[ (2 \sin \theta - \cos \theta)^2 = x^2 \]
Expanding this using the identity \( (A - B)^2 = A^2 + B^2 - 2AB \):
\[ 4 \sin^2 \theta + \cos^2 \theta - 4 \sin \theta \cos \theta = x^2 \]

### Step 5: Add the two equations
Now we will add the two equations we have derived:
1. \( \sin^2 \theta + 4 \cos^2 \theta + 4 \sin \theta \cos \theta = 1 \)
2. \( 4 \sin^2 \theta + \cos^2 \theta - 4 \sin \theta \cos \theta = x^2 \)

Adding these gives:
\[ (\sin^2 \theta + 4 \sin^2 \theta) + (4 \cos^2 \theta + \cos^2 \theta) + (4 \sin \theta \cos \theta - 4 \sin \theta \cos \theta) = 1 + x^2 \]
This simplifies to:
\[ 5 \sin^2 \theta + 5 \cos^2 \theta = 1 + x^2 \]

### Step 6: Use the Pythagorean identity
We know that \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ 5(1) = 1 + x^2 \]
This simplifies to:
\[ 5 = 1 + x^2 \]

### Step 7: Solve for \( x^2 \)
Rearranging gives:
\[ x^2 = 5 - 1 = 4 \]

### Step 8: Find \( x \)
Taking the square root of both sides:
\[ x = \sqrt{4} = 2 \]

### Conclusion
Thus, we find that:
\[ 2 \sin \theta - \cos \theta = 2 \]