Find all \( x \)-values where the function \( g(x)=-7 x+7 \sin (2 x) \) has a horizontal tangent line. S

To find the \( x \)-values where the function \( g(x) = -7x + 7\sin(2x) \) 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 \( g(x) \) with respect to \( x \) is given by: \[ g'(x) = -7 + 14\cos(2x) \] To find the values of \( x \) where the function has a horizontal tangent line, we set the derivative equal to zero and solve for \( x \): \[ -7 + 14\cos(2x) = 0 \] Now, we can solve this equation to find the values of \( x \) where the function has a horizontal tangent line. Solve the equation by following steps: - step0: Solve for \(x\): \(-7+14\cos\left(2x\right)=0\) - step1: Move the constant to the right side: \(14\cos\left(2x\right)=0+7\) - step2: Remove 0: \(14\cos\left(2x\right)=7\) - step3: Divide both sides: \(\frac{14\cos\left(2x\right)}{14}=\frac{7}{14}\) - step4: Divide the numbers: \(\cos\left(2x\right)=\frac{1}{2}\) - step5: Use the inverse trigonometric function: \(2x=\arccos\left(\frac{1}{2}\right)\) - step6: Calculate: \(\begin{align}&2x=\frac{\pi }{3}\\&2x=\frac{5\pi }{3}\end{align}\) - step7: Add the period: \(\begin{align}&2x=\frac{\pi }{3}+2k\pi ,k \in \mathbb{Z}\\&2x=\frac{5\pi }{3}+2k\pi ,k \in \mathbb{Z}\end{align}\) - step8: Calculate: \(\begin{align}&x=\frac{\pi }{6}+k\pi ,k \in \mathbb{Z}\\&2x=\frac{5\pi }{3}+2k\pi ,k \in \mathbb{Z}\end{align}\) - step9: Calculate: \(\begin{align}&x=\frac{\pi }{6}+k\pi ,k \in \mathbb{Z}\\&x=\frac{5\pi }{6}+k\pi ,k \in \mathbb{Z}\end{align}\) - step10: Find the union: \(x=\left\{ \begin{array}{l}\frac{\pi }{6}+k\pi \\\frac{5\pi }{6}+k\pi \end{array}\right.,k \in \mathbb{Z}\) The values of \( x \) where the function \( g(x) = -7x + 7\sin(2x) \) has a horizontal tangent line are given by: \[ x = \frac{\pi}{6} + k\pi \text{ or } x = \frac{5\pi}{6} + k\pi \] where \( k \) is an integer.