Solve \( \mathrm{k}=\frac{1+\sin \mathrm{x}}{\mathrm{n}} \)

To solve \( k = \frac{1 + \sin x}{n} \) for \( x \), follow these steps: 1. Multiply both sides by \( n \): \[ k \cdot n = 1 + \sin x \] 2. Subtract 1 from both sides: \[ \sin x = k \cdot n - 1 \] 3. Take the inverse sine of both sides: \[ x = \arcsin(k \cdot n - 1) + 2\pi m \quad \text{or} \quad x = \pi - \arcsin(k \cdot n - 1) + 2\pi m \] where \( m \) is any integer. **Condition:** \( -1 \leq k \cdot n - 1 \leq 1 \) for real solutions. So, the solutions are: \[ x = \arcsin(k \cdot n - 1) + 2\pi m \quad \text{or} \quad x = \pi - \arcsin(k \cdot n - 1) + 2\pi m \] for any integer \( m \).