sin (2x-45°)=sin 15°,-360°< x < 180°

To solve the given equation, we need to use the inverse sine function. First, let's calculate the reference angle for sin 15°: The reference angle is found by subtracting the given angle from 90°: Reference angle = 90° - 15° = 75° Since sin 15° is positive, the angle lies in the first and second quadrants. So, there are two solutions for (2x - 45°) = 15°: Solution 1: 2x - 45° = 15° 2x = 15° + 45° 2x = 60° x = 60°/2 x = 30° Solution 2: We now consider the second solution in the second quadrant: 2x - 45° = 180° - 15° (180° - 15° because it is the supplementary angle in the second quadrant) 2x - 45° = 165° 2x = 165° + 45° 2x = 210° x = 210°/2 x = 105° Therefore, the solutions to the equation sin (2x - 45°) = sin 15°, in the given range of -360° < x < 180°, are x = 30° and x = 105°.