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

To solve the equation \(\sin 2x = \cos(x + 60^\circ)\), we can use the co-function identity, which states that \(\sin \theta = \cos(90^\circ - \theta)\). First, we rewrite \(\cos(x + 60^\circ)\) in terms of sine: \[ \sin 2x = \cos(x + 60^\circ) \implies \sin 2x = \sin(90^\circ - (x + 60^\circ)) \] Thus, we have: \[ \sin 2x = \sin(30^\circ - x) \] Now, we can set up the general cases for solving \(\sin A = \sin B\): 1. \(2x = 30^\circ - x + 360^\circ n\) 2. \(2x = 180^\circ - (30^\circ - x) + 360^\circ n\) where \(n\) is any integer. **For the first case:** \[ 2x = 30^\circ - x + 360^\circ n \implies 3x = 30^\circ + 360^\circ n \implies x = 10^\circ + 120^\circ n \] **For the second case:** \[ 2x = 180^\circ - (30^\circ - x) + 360^\circ n \implies 2x = 180^\circ - 30^\circ + x + 360^\circ n \] \[ 2x = 150^\circ + x + 360^\circ n \implies x = 150^\circ + 360^\circ n \] Now we compile our solutions: 1. \(x = 10^\circ + 120^\circ n\) 2. \(x = 150^\circ + 360^\circ n\) Combining both solutions, the general solutions of the given equation \(\sin 2x = \cos(x + 60^\circ)\) are: \[ x = 10^\circ + 120^\circ n \quad \text{and} \quad x = 150^\circ + 360^\circ n \] where \(n\) is any integer.