Pattern 2 The common factor Solve for \( A \) if \( A \in\left(-180^{\circ} ; 180^{\circ}\right) \) \[ \sin ^{2} A-\sin A \cos A=0 \]

We start with the equation \[ \sin^2 A - \sin A \cos A = 0. \] ### Step 1. Factor the Equation Notice that both terms contain \(\sin A\). Factor it out: \[ \sin A \left(\sin A - \cos A\right) = 0. \] ### Step 2. Set Each Factor Equal to Zero For the product to be zero, we need either \[ \sin A = 0 \quad \text{or} \quad \sin A - \cos A = 0. \] #### Case 1: \(\sin A = 0\) The general solution for \(\sin A = 0\) is \[ A = 180^\circ \cdot k, \quad k \in \mathbb{Z}. \] Since \(A \in (-180^\circ, 180^\circ)\), the only valid solution from this family is \[ A = 0^\circ. \] #### Case 2: \(\sin A - \cos A = 0\) Rewrite the equation: \[ \sin A = \cos A. \] Dividing both sides by \(\cos A\) (assuming \(\cos A \neq 0\)) gives \[ \tan A = 1. \] The general solution for \(\tan A = 1\) is \[ A = 45^\circ + 180^\circ \cdot k, \quad k \in \mathbb{Z}. \] Within the interval \( (-180^\circ, 180^\circ) \), we obtain: - For \(k = 0\): \(A = 45^\circ\). - For \(k = -1\): \(A = 45^\circ - 180^\circ = -135^\circ\). ### Final Answer The solutions for \( A \in (-180^\circ, 180^\circ) \) are: \[ A = -135^\circ,\quad 0^\circ,\quad 45^\circ. \]