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, we start by rearranging it to isolate the sine and cosine terms: \[ 5\sin(x) = 12\cos(x) \] Dividing both sides by \( \cos(x) \) (assuming \( \cos(x) \neq 0 \)), we obtain: \[ \tan(x) = \frac{12}{5} \] Next, to find the angle whose tangent is \( \frac{12}{5} \), we can let \( r = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \). Now we set values for \( \sin \) and \( \cos \): \[ \sin(x) = \frac{12}{13}, \quad \cos(x) = \frac{5}{13} \] Using the identity \( \sin(x) = R\sin(x + \phi) \) where \( R = 13 \) and \( \tan(\phi) = \frac{12}{5} \): Now we find \( \phi \) using the tangent: \[ \phi = \tan^{-1}\left(\frac{12}{5}\right) \] Now, rewrite the original equation in the form: \[ 13\sin(x + \phi) = 0 \implies \sin(x + \phi) = 0 \] The solutions for \( \sin(x + \phi) = 0 \) are given by: \[ x + \phi = n\pi \] Thus, \[ x = n\pi - \phi \] Considering \( n = 0 \) and \( n = 1 \) for one period \( [0, 2\pi) \): 1. \( x = -\phi \) (not within \([0, 2\pi)\)) 2. \( x = \pi - \phi \) So we compute \( \phi \): \[ \phi = \tan^{-1}\left(\frac{12}{5}\right) \approx 1.176 \] Calculating \( \pi - \phi \): \[ \pi - \phi \approx 3.142 - 1.176 \approx 1.966 \] To summarize, the solutions in the interval \( [0, 2\pi) \) are: \[ x \approx 1.966 \quad \text{and} \quad x = \pi + (n\pi - \phi) \text{ for } n = 1 \rightarrow 4.142 \] Thus the possible values of \( x \) are approximately: \[ 1.966 \text{ and } 4.142 \]