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

To find the \( x \)-values where the function \( f(x)=-\frac{3}{2} x+3 \sin x \) has a horizontal tangent line, we need to find the values of \( x \) for which the derivative of the function is equal to zero. The derivative of the function \( f(x)=-\frac{3}{2} x+3 \sin x \) with respect to \( x \) is given by: \[ f'(x) = -\frac{3}{2} + 3 \cos x \] To find the values of \( x \) for which the derivative is equal to zero, we set \( f'(x) = 0 \) and solve for \( x \): \[ -\frac{3}{2} + 3 \cos x = 0 \] Solving this equation will give us the \( x \)-values where the function has a horizontal tangent line. Solve the equation by following steps: - step0: Solve for \(x\): \(-\frac{3}{2}+3\cos\left(x\right)=0\) - step1: Move the constant to the right side: \(3\cos\left(x\right)=0+\frac{3}{2}\) - step2: Remove 0: \(3\cos\left(x\right)=\frac{3}{2}\) - step3: Multiply by the reciprocal: \(3\cos\left(x\right)\times \frac{1}{3}=\frac{3}{2}\times \frac{1}{3}\) - step4: Multiply: \(\cos\left(x\right)=\frac{1}{2}\) - step5: Use the inverse trigonometric function: \(x=\arccos\left(\frac{1}{2}\right)\) - step6: Calculate: \(\begin{align}&x=\frac{\pi }{3}\\&x=\frac{5\pi }{3}\end{align}\) - step7: Add the period: \(\begin{align}&x=\frac{\pi }{3}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{5\pi }{3}+2k\pi ,k \in \mathbb{Z}\end{align}\) - step8: Find the union: \(x=\left\{ \begin{array}{l}\frac{\pi }{3}+2k\pi \\\frac{5\pi }{3}+2k\pi \end{array}\right.,k \in \mathbb{Z}\) The \( x \)-values where the function \( f(x)=-\frac{3}{2} x+3 \sin x \) has a horizontal tangent line are given by: \[ x = \frac{\pi}{3} + 2k\pi \text{ or } x = \frac{5\pi}{3} + 2k\pi \] where \( k \) is an integer.