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 the Equation in the Form \( R\sin(x + \alpha) = 0 \)** The general form for such an expression is: \[ a\sin(x) + b\cos(x) = R\sin(x + \alpha) \] where: \[ R = \sqrt{a^2 + b^2} \] \[ \alpha = \arctan\left(\frac{b}{a}\right) \] For the given equation: \[ a = 5 \quad \text{and} \quad b = -12 \] Thus, \[ R = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \] \[ \alpha = \arctan\left(\frac{-12}{5}\right) = -\arctan\left(\frac{12}{5}\right) \] ### **2. Rewrite the Original Equation** \[ 5\sin(x) - 12\cos(x) = 13\sin\left(x - \arctan\left(\frac{12}{5}\right)\right) = 0 \] ### **3. Solve for \( x \)** Set the sine function to zero: \[ 13\sin\left(x - \arctan\left(\frac{12}{5}\right)\right) = 0 \implies \sin\left(x - \arctan\left(\frac{12}{5}\right)\right) = 0 \] \[ x - \arctan\left(\frac{12}{5}\right) = k\pi \quad \text{for integer } k \] \[ \Rightarrow x = \arctan\left(\frac{12}{5}\right) + k\pi \] ### **4. Find All Solutions Within One Period (\( 0 \leq x < 2\pi \))** The principal solutions within one period are: \[ x = \arctan\left(\frac{12}{5}\right) \quad \text{and} \quad x = \arctan\left(\frac{12}{5}\right) + \pi \] ### **Final Answer** All real numbers congruent to arctan (12⁄5) modulo π. In other words, the solutions are x = arctan(12/5) + k π for all integers k.