10Solving 2sinx−1=0
Solve 2 sin x − 1 = 0 for 0° ≤ x < 360°.
Isolating gives sin x = 1/2, and since sine is positive in quadrants I and II with reference angle 30°, the solutions are x = 30° and x = 150°. Check: 2(1/2) − 1 = 0 ✓.
Solve 2 sin x − 1 = 0 for 0° ≤ x < 360°.
Isolating gives sin x = 1/2, and since sine is positive in quadrants I and II with reference angle 30°, the solutions are x = 30° and x = 150°. Check: 2(1/2) − 1 = 0 ✓.