Trig

Find all solutions of the equation in the interval [0,2pi)

2 cos^2 x-cos x = 0

-2cos^2 + cosx + 0
(x+1/2) (x+0/2)
(2x+1) (x+0)
-1/2,0
2Pi/3, 4pi/3, pi/2, 3pi/2

my teacher circled pi/2 and 3pi/2

What did I do wrong? I don't understand...

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  1. Hard to figure out what you are doing.

    2 cos^2 x-cos x = 0
    cosx(2cosx - 1) = 0
    cosx = 0 or cosx = 1/2

    if cosx = 0,
    x = π/2 or x = 3π/2

    if cosx = 1/2
    x = π/3 or x = 2π - π/3 = 5π/3

    so you have 4 answers:
    π/2, 3π/2, π/3, and 5π/3

    (in degrees: 90°, 270°, 60°, 300° , they all work)

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    posted by Reiny

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