How many solutions does the equation 2 sin^2 x + 3 cos x - 3 = 0 have in the interval [0, 2π)?
Answer and explanation
Correct answer: B. 3
- Step 1: Replace sin^2 x with 1 - cos^2 x: 2 - 2cos^2 x + 3 cos x - 3 = 0, which simplifies to 2cos^2 x - 3 cos x + 1 = 0.
- Step 2: Factor: (2 cos x - 1)(cos x - 1) = 0, so cos x = 1/2 or cos x = 1.
- Step 3: In [0, 2π): cos x = 1 gives x = 0; cos x = 1/2 gives x = π/3 and 5π/3. That is 3 solutions.
Answer
3.
Why not the others: '2' leaves out x = 0; '4' counts x = 2π as well, which is excluded from the interval; '5' counts the root x = 0 more than once.
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