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Solve the given equation in the interval [0,2 pi].
Note: The answer must be written as a multiple of pi. Give exact
answers. Do not use decimal numbers. The answer must
an integer or a fraction. Note that pi is already provided with the
answer so you just have to find the appropriate multiple. E.g.
the answer is p/2 you should enter 1/2. If there is more than one
answer write them separated by commas.
2(sin(x))^2−5cos(x)+1 = 0
x=__________ pi

  • Math -

    2(sin(x))^2−5cos(x)+1 = 0
    2(1- cos^2(x))−5cos(x)+1 = 0
    2 - 2cos^2(x) - 5cos(x) + 1 = 0
    2cos^2(x) + 5cos(x) + 3 = 0

    (cosx + 1 )(2cosx + 3) = 0
    cosx = -1 or cosx = -3/2

    from the first part x = 0 or 2pi
    there is no solution to the second part since the cosine function cannot exceed ±1

  • Math -

    Thats not correct....
    I'm not sure how you got from
    2 - 2cos^2(x) - 5cos(x) + 1 = 0
    2cos^2(x) + 5cos(x) + 3 = 0
    It should be -2cos^2(x) - 5cos(x)+1=0

    From there I'm stuck.

  • Math -

    sorry. meant
    -2cos^2(x) - 5cos(x)+3=0

  • Math -

    Oh my, OH my, two errors in the same post, I must be getting careless.

    Let's start from

    2(1- cos^2(x))−5cos(x)+1 = 0
    2 - 2cos^2(x) - 5cos(x) + 1 = 0
    2cos^2(x) + 5cos(x) - 3 = 0
    (cosx + 3)(2cosx - 1) = 0
    cosx = -3 which will not produce an answer
    cosx = 1/2
    so x = pi/3 or 5pi/3 (60º or 300º)

    (these answers are correct, I tested them)

  • Math -

    Thank you....the pi/3, 5pi/3 is correct.

  • Math -

    Thanks for catching my error,
    of course if cos x = -1 then x = pi.

    I read my own writing as cosx = 1, (let's blame it on eyesight of an old guy, lol)

    you said:"I'm not sure how you got from
    2 - 2cos^2(x) - 5cos(x) + 1 = 0 "

    look at 2(1- cos^2(x)) and expand it, you get
    2 - 2cos^x

    so x = pi is your only solution

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