Solve for `theta(0^(@) lt theta lt 90^(@))` :
`sin^(2)theta-1/2sintheta=0`

Answer

To solve the equation \( \sin^2 \theta - \frac{1}{2} \sin \theta = 0 \) for \( \theta \) in the range \( 0^\circ < \theta < 90^\circ \), follow these steps:

### Step 1: Rewrite the Equation
Start with the original equation:
\[ \sin^2 \theta - \frac{1}{2} \sin \theta = 0 \]

### Step 2: Factor the Equation
We can factor out \( \sin \theta \) from the equation:
\[ \sin \theta \left( \sin \theta - \frac{1}{2} \right) = 0 \]

### Step 3: Set Each Factor to Zero
Now, we have two factors. Set each factor equal to zero:
1. \( \sin \theta = 0 \)
2. \( \sin \theta - \frac{1}{2} = 0 \) which simplifies to \( \sin \theta = \frac{1}{2} \)

### Step 4: Solve Each Equation
1. For \( \sin \theta = 0 \):
- The solutions are \( \theta = 0^\circ \) or \( \theta = 180^\circ \). However, since \( \theta \) must be in the range \( 0^\circ < \theta < 90^\circ \), this solution is not valid.

2. For \( \sin \theta = \frac{1}{2} \):
- The angle \( \theta \) that satisfies this equation in the range \( 0^\circ < \theta < 90^\circ \) is \( \theta = 30^\circ \).

### Step 5: Conclusion
Thus, the only valid solution for \( \theta \) in the specified range is:
\[ \theta = 30^\circ \]