Find all \( x \)-values where the function \( f(x)=-\frac{3}{2} x+3 \sin x \) has a horizontal tangent line

To find the values of \( x \) where the function \( f(x) = -\frac{3}{2} x + 3 \sin x \) has a horizontal tangent line, we first need to calculate the derivative \( f'(x) \) and set it equal to zero. The derivative is calculated as follows: \[ f'(x) = -\frac{3}{2} + 3 \cos x \] Next, we set the derivative equal to zero to find the points where the tangent line is horizontal: \[ -\frac{3}{2} + 3 \cos x = 0 \] Solving for \( \cos x \): \[ 3 \cos x = \frac{3}{2} \] \[ \cos x = \frac{1}{2} \] The cosine function equals \( \frac{1}{2} \) at the following angles: \[ x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad x = -\frac{\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \] Thus, the \( x \)-values where the function \( f(x) \) has a horizontal tangent line are: \[ x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad x = -\frac{\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}). \]