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

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Weston Evans · Accepted Answer

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 $$ m $$.

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

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

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