Given the equation \( 5\sin(x) - 12\cos(x) = 0 \), use the auxiliary angle method to find all possible values of x within one period.

To solve the equation \( 5\sin(x) - 12\cos(x) = 0 \) using the **auxiliary angle method**, follow these steps: 1. **Express in the Form \( R\sin(x + \alpha) \):** The expression \( 5\sin(x) - 12\cos(x) \) can be rewritten using the auxiliary angle formula: \[ a\sin(x) + b\cos(x) = R\sin(x + \alpha) \] where: \[ R = \sqrt{a^2 + b^2} \] \[ \tan(\alpha) = \frac{b}{a} \] For the given equation: \[ a = 5, \quad b = -12 \] \[ R = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \] \[ \tan(\alpha) = \frac{-12}{5} \implies \alpha = \arctan\left(-\frac{12}{5}\right) = -\arctan\left(\frac{12}{5}\right) \] Thus, the equation becomes: \[ 13\sin(x - \arctan(12/5)) = 0 \] Simplifying: \[ \sin\left(x - \arctan\left(\frac{12}{5}\right)\right) = 0 \] 2. **Solve for \( x \):** The sine function equals zero at integer multiples of \( \pi \): \[ x - \arctan\left(\frac{12}{5}\right) = n\pi \quad \text{where } n \text{ is an integer} \] Solving for \( x \): \[ x = n\pi + \arctan\left(\frac{12}{5}\right) \] Within one period (\( 0 \leq x < 2\pi \)), there are two solutions corresponding to \( n = 0 \) and \( n = 1 \): \[ x_1 = \arctan\left(\frac{12}{5}\right) \] \[ x_2 = \pi + \arctan\left(\frac{12}{5}\right) \] 3. **Final Answer:** All solutions within one period are: \[ x = \arctan\left(\frac{12}{5}\right) \quad \text{and} \quad x = \pi + \arctan\left(\frac{12}{5}\right) \] **Summary:** The solutions to \( 5\sin(x) - 12\cos(x) = 0 \) within one period \( [0, 2\pi) \) are: \[ x = \arctan\left(\frac{12}{5}\right) \quad \text{and} \quad x = \pi + \arctan\left(\frac{12}{5}\right) \] Answer: All real numbers congruent to arctan 12⁄5 or to π plus arctan 12⁄5. Thus, within one period, x = arctan(12/5) and x = π + arctan(12/5)