Solving a Trigonometric Equation on a Restricted Domain
A student is solving 2sin²x + 3cos x − 3 = 0 for 0 ≤ x ≤ 2π. They convert sin²x to 1 − cos²x and reach 2 − 2cos²x + 3cos x − 3 = 0, which simplifies to 2cos²x − 3cos x + 1 = 0. They factor this as (2cos x − 1)(cos x − 1) = 0, obtaining cos x = 1/2 or cos x = 1. They then write 'x = π/3 or x = 0' and submit. A marker notices the answer is incomplete. The student protests that the quadratic gave exactly two roots, so there should be two solutions. The marker must explain where the missing solutions arise and whether any solutions could be extraneous.
1. Identify the student's error and give the complete solution set on 0 ≤ x ≤ 2π.
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The algebra is correct; the error is in solving the trigonometric equations. Each value of cos x on the unit circle generally corresponds to two angles in [0, 2π] (except at the endpoints 0 and 2π, and at π).
For cos x = 1/2: x = π/3 or x = 2π − π/3 = 5π/3. For cos x = 1: x = 0 (and x = 2π if the closed interval endpoint is counted).
So the full solution set is x = 0, π/3, 5π/3, and also 2π if endpoints included. The student gave only one angle per cosine value, losing 5π/3 (and 2π). The number of roots of the quadratic (two) is not the number of solutions for x; the quadratic counts distinct cosine values, not angles.
2. Explain whether any of the solutions could be extraneous, and verify one solution numerically.
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No extraneous solutions arise here. The substitution sin²x = 1 − cos²x is an identity true for all real x, so it does not restrict or enlarge the domain; every x satisfying the original equation satisfies the cosine equation and conversely. Extraneous roots would appear only if we had squared both sides or multiplied by an expression that could be zero.
Check x = 5π/3: cos(5π/3) = 1/2, sin(5π/3) = −√3/2, so sin²(5π/3) = 3/4. Then 2(3/4) + 3(1/2) − 3 = 1.5 + 1.5 − 3 = 0. Confirmed. Check x = 0: 2(0) + 3(1) − 3 = 0. Confirmed. Both satisfy the original equation, so all listed solutions are genuine.