`cos theta + sin theta - sin 2 theta = (1)/(2), 0 lt theta lt (pi)/(2)`

Answer

To solve the equation \( \cos \theta + \sin \theta - \sin 2\theta = \frac{1}{2} \) for \( 0 < \theta < \frac{\pi}{2} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the original equation:
\[ \cos \theta + \sin \theta - \sin 2\theta = \frac{1}{2} \]
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Substituting this into the equation gives:
\[ \cos \theta + \sin \theta - 2 \sin \theta \cos \theta = \frac{1}{2} \]

### Step 2: Rearranging the equation
Rearranging the equation, we have:
\[ \cos \theta + \sin \theta - 2 \sin \theta \cos \theta - \frac{1}{2} = 0 \]
This can be rearranged to:
\[ \sin \theta - 2 \sin \theta \cos \theta = \frac{1}{2} - \cos \theta \]

### Step 3: Factor out common terms
Now, we can factor out \( \sin \theta \) from the left-hand side:
\[ \sin \theta (1 - 2 \cos \theta) = \frac{1 - 2 \cos \theta}{2} \]

### Step 4: Set the equation to zero
We can simplify this to:
\[ 2 \sin \theta (1 - 2 \cos \theta) - (1 - 2 \cos \theta) = 0 \]
Factoring out \( (1 - 2 \cos \theta) \):
\[ (1 - 2 \cos \theta)(2 \sin \theta - 1) = 0 \]

### Step 5: Solve each factor
Now we have two factors to solve:
1. \( 1 - 2 \cos \theta = 0 \)
2. \( 2 \sin \theta - 1 = 0 \)

#### Factor 1: Solve \( 1 - 2 \cos \theta = 0 \)
\[ 2 \cos \theta = 1 \implies \cos \theta = \frac{1}{2} \]
This gives:
\[ \theta = \frac{\pi}{3} \]

#### Factor 2: Solve \( 2 \sin \theta - 1 = 0 \)
\[ 2 \sin \theta = 1 \implies \sin \theta = \frac{1}{2} \]
This gives:
\[ \theta = \frac{\pi}{6} \]

### Step 6: Consider the range
Since \( 0 < \theta < \frac{\pi}{2} \), both solutions \( \theta = \frac{\pi}{3} \) and \( \theta = \frac{\pi}{6} \) are valid.

### Final Answer
Thus, the solutions to the equation are:
\[ \theta = \frac{\pi}{3} \quad \text{and} \quad \theta = \frac{\pi}{6} \]

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