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Here is the problem: Find arcsec(negative 2) and every angle in [0°, 360°) with sec x equals negative 2.

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Find the angle whose sec is negative 2: the principal value of arcsec, every solution of sec(x) equals negative 2 in one turn, and the general solution.

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Rewrite sec as a reciprocal.

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Principal value of arcsec, taken in the range [0, pi], excluding pi/2. 120° is a special angle, so the answer is exact.

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Cosine is even, so x and negative x (that is, 360° minus x) give the same value.

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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 arcsec(negative 2) equals 120° (2pi/3); solutions in [0°, 360°): 120°, 240°; general: x equals plus or minus 120° plus 360°n.

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

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Substituting each solution back: sec(120°) approximately negative 2; sec(240°) approximately negative 2.