Question

2 - 2 cos² θ°= sin θ° , 0° ≤ θ ≤ 360°

Answer 1

2 - 2 cos² x = sin x
2 = sin x + 2 cos² x
0 = -2 + sin x + 2(1 - sin² x)
0 = -2 sin² x + sin x + 2 - 2
0 = 2 sin² x - sin x
0 = sin x (2 sin x - 1)

sin x = 0 v sin x = 0.5
x = 0° v 30°

x1 = 0° + k.360°
x1 = 0°, 360°

x2 = (180° - 0°) + k.360°
x2 = 180°

x3 = 30° + k.360°
x3 = 30°

x4 = (180° - 30°) + k.360°
x4 = 150°

HP: {x | 0°, 30°, 150°, 180°, 360°}

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