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Here is the problem: Find arcsin(1 over 2) and every angle in [0°, 360°) with sin x equals 1 over 2.

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Find the angle whose sin is 1 over 2: the principal value of arcsin, every solution of sin(x) equals 1 over 2 in one turn, and the general solution.

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Principal value of arcsin, taken in the range [−pi/2, pi/2]. 30° is a special angle, so the answer is exact.

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Sine is also positive or negative in a second quadrant, so use x and 180° minus x.

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Solutions in one full turn, 0° less than or equal to x less than 360°.

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General solution, n any integer.

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So the answer is arcsin(1 over 2) equals 30° (pi/6); solutions in [0°, 360°): 30°, 150°; general: x equals 30° plus 360°n or x equals 150° plus 360°n.

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An inverse trigonometric function returns one angle, the principal value. Because the trigonometric functions repeat, the equation sin(x) equals 1 over 2 has infinitely many solutions; the calculator lists those in one turn and the general pattern.

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Substituting each solution back: sin(30°) approximately 0.5; sin(150°) approximately 0.5.