Determine the general solution of: \[ \sin 2 x=\cos (x+60) \]

Solve the equation \( \sin(2x)=\cos(x+60) \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sin\left(2x\right)=\cos\left(x+60\right)\) - step1: Move the expression to the left side: \(\sin\left(2x\right)-\cos\left(x+60\right)=0\) - step2: Transform the expression: \(-2\sin\left(\frac{\pi +120-2x}{4}\right)\sin\left(\frac{\pi -120-6x}{4}\right)=0\) - step3: Elimination the left coefficient: \(\sin\left(\frac{\pi +120-2x}{4}\right)\sin\left(\frac{\pi -120-6x}{4}\right)=0\) - step4: Separate into possible cases: \(\begin{align}&\sin\left(\frac{\pi +120-2x}{4}\right)=0\\&\sin\left(\frac{\pi -120-6x}{4}\right)=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=\frac{\pi +120}{2}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{41\pi -120}{6}+\frac{2k\pi }{3},k \in \mathbb{Z}\end{align}\) - step6: Find the union: \(x=\left\{ \begin{array}{l}\frac{41\pi -120}{6}+\frac{2k\pi }{3}\\\frac{\pi +120}{2}+2k\pi \end{array}\right.,k \in \mathbb{Z}\) The general solution of the equation \(\sin 2x = \cos (x+60)\) is given by: \[ x = \left\{ \begin{array}{l} \frac{41\pi -120}{6} + \frac{2k\pi}{3} \\ \frac{\pi +120}{2} + 2k\pi \end{array} \right., k \in \mathbb{Z} \] where \(k\) is an integer.