Solve the trigonometric equation 2 cos^2(x) - cos(x) - 1 = 0 on the interval [0, 2pi).

Answer and explanation

Correct answer: C. x = 0, 2 pi / 3, 4 pi / 3

  1. Step 1: Treat as quadratic in u = cos(x): 2u^2 - u - 1 = 0.
  2. Step 2: Factor: (2u + 1)(u - 1) = 0.
  3. Step 3: Roots: u = 1 or u = -1/2.
  4. Step 4: Case 1: cos(x) = 1 on [0, 2pi) => x = 0.
  5. Step 5: Case 2: cos(x) = -1/2 in Quadrants II and III => x = pi - pi/3 = 2pi/3, and x = pi + pi/3 = 4pi/3.
  6. Step 6: Assemble solution set: x = 0, 2pi/3, 4pi/3.
Answer
x = 0, 2 pi / 3, 4 pi / 3.
Why not the others: 'x = pi / 3, pi, 5 pi / 3' solves for cos(x) = 1/2 and cos(x) = -1; 'x = 0, pi / 3, 5 pi / 3' confuses signs of cosine roots; 'x = pi / 2, 3 pi / 2' solves for sin(x) = +-1.

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