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Jelillia J
Jelillia J asked in Science & MathematicsMathematics · 10 years ago

0=2cos(2x)-cos(x)....find x?

identities and factoring of some sort. I tried for about an hour to get this and couldn't. please help

3 Answers

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  • No Mythology
    Lv 7
    10 years ago

    cos(2x) = 2cos²(x) - 1. So

    2cos(2x) - cos(x) = 4cos²(x) - 2 - cos(x) = 0 ==>

    cos(x) = 1/8 ± √(33)/8

    Both of these numbers are between -1 and 1. To get all possible solutions

    x = ± arccos((1 + √(33))/8) + 2nπ or x = ± arccos((1 - √(33))/8) + 2nπ

    n any integer.

    The first representation gives all quadrant I and IV solution and the second gives all quadrant II and III solutions.

  • Mike G
    Lv 7
    10 years ago

    0=2[cos^2(x) - sin^2(x)]-cos(x)

    0=2cos^2(x)-2sin^2(x)-cos(x)

    0=2cos^2(x) - 2[1-cos^2(x)] -cos(x)

    0=4cos^2(x)-cos(x)-2

    Use the quadratic formula

    cos(x) = 0.843 or -0.593

    x = 32.54, 327.46, 126.37 and 233.63 degrees

  • Andrew Baxton
    10 years ago

    http://www.wolframalpha.com/input/?i=0%3D2cos%282x...

    Source(s): Math Student
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