4.1 If \( 13 \sin \theta+5=0 \) and \( \tan \theta>0 \), determine \( 3 \cos \theta \)

To solve the equation \( 13 \sin \theta + 5 = 0 \), we first isolate \( \sin \theta \): \[ \sin \theta = -\frac{5}{13} \] Since \( \tan \theta > 0 \), \( \theta \) must be in either the first or third quadrant. However, \( \sin \theta \) is negative, so \( \theta \) must be in the third quadrant. Next, we can find \( \cos \theta \) using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting the value of \( \sin \theta \): \[ \left(-\frac{5}{13}\right)^2 + \cos^2 \theta = 1 \] \[ \frac{25}{169} + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - \frac{25}{169} = \frac{169}{169} - \frac{25}{169} = \frac{144}{169} \] Taking the square root gives: \[ \cos \theta = -\frac{12}{13} \] We take the negative root, since \(\theta\) is in the third quadrant. Now, we can find \( 3 \cos \theta \): \[ 3 \cos \theta = 3 \left(-\frac{12}{13}\right) = -\frac{36}{13} \] Thus, the value of \( 3 \cos \theta \) is \[ \boxed{-\frac{36}{13}} \]