trig

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solve for x if x is in radians.

cos2(x) + 4cos(x) = -3

  • trig -

    Is that
    cos^2 x + 4cosx = -3 ??
    if so, then let cosx = y to get
    y^2 + 4y + 3 = 0
    (y+1)(y+3) = 0
    y = -1 or y = -3
    so
    cosx = -1 ---> x = π
    or
    cosx = -3 , not possible

    so x = π

  • trig -

    no, its not squared. it is (cosx)(2)

  • trig -

    (cosx)(2) wouldn't make much sense, then it would be
    2cosx and the entire equation would be
    6cosx = -3
    cosx = -1/2
    and x = 120° or 240°

    did you mean cos(2x) ?
    then you can use the identity
    cos(2x) = 2cos^2x - 1

    cos((x) + 4cos(x) = -3
    2cos^2x - 1 + 4cosx = -3
    2cos^2x + 4cosx + 2 = 0
    cos^2x + 2cosx + 1 = 0
    (cosx + 1)^2 = 0
    cosx + 1 = 0
    cosx = -1
    x = 180° or x = π
    (which by coincidence is the same answer to the other interpretation)

  • trig -

    okay i think that is right. thank you

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