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

To solve the equation \( \sin^2 A - \sin A \cos A = 0 \), we can factor out a common term: \[ \sin A (\sin A - \cos A) = 0 \] This gives us two possible cases to consider: 1. \( \sin A = 0 \) 2. \( \sin A - \cos A = 0 \) **Case 1: \( \sin A = 0 \)** The solutions occur at: \[ A = 0^{\circ}, \, 180^{\circ}, \, -180^{\circ} \] In the interval \( (-180^{\circ}, 180^{\circ}) \), we have: \[ A = 0^{\circ} \] **Case 2: \( \sin A - \cos A = 0 \)** This can be rewritten as: \[ \sin A = \cos A \] Dividing both sides by \( \cos A \) (where \( \cos A \neq 0 \)) gives: \[ \tan A = 1 \] The solutions for \( A \) when \( \tan A = 1 \) are: \[ A = 45^{\circ} + k \cdot 180^{\circ}, \quad k \in \mathbb{Z} \] Considering \( k = 0 \) and \( k = -1 \): - \( A = 45^{\circ} \) - \( A = -135^{\circ} \) Now, we compile our solutions: From Case 1: - \( A = 0^{\circ} \) From Case 2: - \( A = 45^{\circ} \) - \( A = -135^{\circ} \) Thus, the solutions for \( A \) in the interval \( (-180^{\circ}, 180^{\circ}) \) are: \[ A = 0^{\circ}, \, 45^{\circ}, \, -135^{\circ} \]