\( ( 5.4 ) \quad \cos 3 x = \cos ( x + 20 ^ { \circ } ) \)

To solve the equation \( \cos 3x = \cos (x + 20^\circ) \), we can use the fact that if \( \cos A = \cos B \), then \( A = B + 360^\circ k \) or \( A = -B + 360^\circ k \), where \( k \) is any integer. 1. From the first case, we have: \[ 3x = x + 20^\circ + 360^\circ k \] Simplifying gives: \[ 2x = 20^\circ + 360^\circ k \] \[ x = 10^\circ + 180^\circ k \] 2. From the second case, we have: \[ 3x = - (x + 20^\circ) + 360^\circ k \] Simplifying gives: \[ 3x = -x - 20^\circ + 360^\circ k \] \[ 4x = -20^\circ + 360^\circ k \] \[ x = -5^\circ + 90^\circ k \] Now, we have two families of solutions: 1. \( x = 10^\circ + 180^\circ k \) 2. \( x = -5^\circ + 90^\circ k \) The complete solution set can be generated by substituting different integer values for \( k \). Here, we can find specific solutions such as \( k=0 \) for the first family gives \( x = 10^\circ \) and \( k=0 \) for the second family gives \( x = -5^\circ \). These solutions represent angles where the two cosine expressions are equal.