(4) \( \operatorname{sen}\left(30^{\circ}+p\right)-\cos \left(60^{\circ}+p\right)=\sqrt{3} \operatorname{sen} \beta \)

To solve the equation \[ \operatorname{sen}\left(30^{\circ}+p\right)-\cos \left(60^{\circ}+p\right)=\sqrt{3} \operatorname{sen} \beta, \] we start by simplifying the left-hand side. 1. **Use trigonometric identities**: - Recall that \(\cos(60^\circ) = \frac{1}{2}\). - Therefore, \(\cos(60^\circ + p) = \cos(60^\circ)\cos(p) - \sin(60^\circ)\sin(p) = \frac{1}{2}\cos(p) - \frac{\sqrt{3}}{2}\sin(p)\). 2. **Substituting into the equation**: The left-hand side becomes: \[ \operatorname{sen}(30^\circ + p) - \cos(60^\circ + p) = \sin(30^\circ)\cos(p) + \cos(30^\circ)\sin(p) - \left(\frac{1}{2}\cos(p) - \frac{\sqrt{3}}{2}\sin(p)\right). \] Since \(\sin(30^\circ) = \frac{1}{2}\) and \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), we have: \[ \frac{1}{2}\cos(p) + \frac{\sqrt{3}}{2}\sin(p) - \left(\frac{1}{2}\cos(p) - \frac{\sqrt{3}}{2}\sin(p)\right). \] 3. **Combine like terms**: \[ \frac{1}{2}\cos(p) + \frac{\sqrt{3}}{2}\sin(p) - \frac{1}{2}\cos(p) + \frac{\sqrt{3}}{2}\sin(p) = \sqrt{3}\sin(p). \] 4. **Setting the equation**: Now, we have: \[ \sqrt{3}\sin(p) = \sqrt{3}\sin(\beta). \] 5. **Dividing both sides by \(\sqrt{3}\)** (assuming \(\sqrt{3} \neq 0\)): \[ \sin(p) = \sin(\beta). \] 6. **Finding the general solution**: The solutions to \(\sin(p) = \sin(\beta)\) are given by: \[ p = \beta + n\cdot 360^\circ \quad \text{or} \quad p = 180^\circ - \beta + n\cdot 360^\circ, \] where \(n\) is any integer. Thus, the final solutions for \(p\) are: \[ p = \beta + n\cdot 360^\circ \quad \text{or} \quad p = 180^\circ - \beta + n\cdot 360^\circ, \quad n \in \mathbb{Z}. \]