\( \sin x \cosh y + i \operatorname { cosin } \sin h y = 0 \)

To solve the equation \[ \sin x \cosh y + i \cos x \sinh y = 0, \] we can separate the real and imaginary parts. The equation can be rewritten as: \[ \sin x \cosh y + i \cos x \sinh y = 0. \] This implies that both the real part and the imaginary part must equal zero: 1. Real part: \[ \sin x \cosh y = 0, \] 2. Imaginary part: \[ \cos x \sinh y = 0. \] ### Solving the Real Part From the equation \(\sin x \cosh y = 0\), we have two cases: - Case 1: \(\sin x = 0\) This occurs when: \[ x = n\pi, \quad n \in \mathbb{Z}. \] - Case 2: \(\cosh y = 0\) However, \(\cosh y\) is never zero for real \(y\) since \(\cosh y \geq 1\) for all real \(y\). Thus, this case does not provide any solutions. ### Solving the Imaginary Part From the equation \(\cos x \sinh y = 0\), we again have two cases: - Case 1: \(\cos x = 0\) This occurs when: \[ x = \frac{\pi}{2} + m\pi, \quad m \in \mathbb{Z}. \] - Case 2: \(\sinh y = 0\) This occurs when: \[ y = 0. \] ### Summary of Solutions Combining the results from both parts, we have: 1. From \(\sin x = 0\): - \(x = n\pi\) for \(n \in \mathbb{Z}\) and \(y\) can be any real number. 2. From \(\cos x = 0\) and \(\sinh y = 0\): - \(x = \frac{\pi}{2} + m\pi\) for \(m \in \mathbb{Z}\) and \(y = 0\). Thus, the complete set of solutions is: - \(x = n\pi\) for \(n \in \mathbb{Z}\) and \(y \in \mathbb{R}\). - \(x = \frac{\pi}{2} + m\pi\) for \(m \in \mathbb{Z}\) and \(y = 0\).