Solve the trigonometric equation 2sin(x)^2 - 3sin(x) + 1 = 0 on the interval [0, 2*pi].
Answer and explanation
Correct answer: C. x = pi/6, 5*pi/6, and pi/2
- Step 1: Factor the quadratic in sin(x): (2*sin(x) - 1)(sin(x) - 1) = 0.
- Step 2: Find roots: sin(x) = 1/2 or sin(x) = 1.
- Step 3: For sin(x) = 1/2 on [0, 2pi], x = pi/6 and 5pi/6. For sin(x) = 1, x = pi/2.
Answer
x = pi/6, 5*pi/6, and pi/2.
Why not the others: pi/3 and 2pi/3 yield sin(x) = sqrt(3)/2; pi/4 and 3pi/4 yield sin(x) = sqrt(2)/2; 7pi/6 and 11pi/6 yield sin(x) = -1/2.
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