2.) Solve for \( \left(0^{\circ}<\theta<90^{\circ}\right) \) 1) \( \sin ^{2} \theta-\frac{1}{2} \sin \theta=0 \)

We are given the equation \[ \sin^2 \theta - \frac{1}{2} \sin \theta = 0 \] to solve for \(0^{\circ} < \theta < 90^{\circ}\). ### Step 1: Factor the Equation Notice that \(\sin \theta\) is common in both terms. Factor it out: \[ \sin \theta \left(\sin \theta - \frac{1}{2}\right) = 0. \] ### Step 2: Set Each Factor Equal to Zero For the product to be zero, at least one of the factors must be zero. 1. **First factor:** \[ \sin \theta = 0 \] 2. **Second factor:** \[ \sin \theta - \frac{1}{2} = 0 \quad \Longrightarrow \quad \sin \theta = \frac{1}{2}. \] ### Step 3: Solve Each Equation Considering the Domain \(0^{\circ} < \theta < 90^{\circ}\) 1. **For \(\sin \theta = 0\):** \(\theta = 0^{\circ}\) or \(\theta = 180^{\circ}\), etc. However, since the interval is \(0^{\circ} < \theta < 90^{\circ}\), \(\theta = 0^{\circ}\) is not included. 2. **For \(\sin \theta = \frac{1}{2}\):** The solution in the interval \(0^{\circ} < \theta < 90^{\circ}\) is \[ \theta = 30^{\circ}. \] ### Final Answer The only valid solution in the interval \(0^{\circ} < \theta < 90^{\circ}\) is: \[ \boxed{30^{\circ}}. \]