WEBVTT

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Here is the problem: Find arccos(negative the square root of 3 over 2 and every angle in [0°, 360°) with cos x equals negative the square root of 3 over 2.

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Find the angle whose cos is negative the square root of 3 over 2: the principal value of arccos, every solution of cos(x) equals negative the square root of 3 over 2 in one turn, and the general solution.

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Principal value of arccos, taken in the range [0, pi]. 150° 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 arccos(negative the square root of 3 over 2 equals 150° (5pi/6); solutions in [0°, 360°): 150°, 210°; general: x equals plus or minus 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 cos(x) equals negative the square root of 3 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: cos(150°) approximately negative 0.8660254038; cos(210°) approximately negative 0.8660254038.