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Math Help - trig equations

  1. #1
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    trig equations

    solve the following equations for all values of x:
    d) 2sin^2(x) = 2 + cosx
    f) cos(-2x) + 2 = 3cos(2x)

    and if it isnt too much to ask T_T
    simplify the following expressions:
    b) cos^2x + tan^2xcos^2x

    thankss
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  2. #2
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    d) 2sin^2(x) = 2 + cosx

    Expand the left hand side:  2(1-cos^2(x))= 2 + cos x
    Move all terms to one side and factorise:  cos(x) (2cos(x)+1) = 0
    I assume you can take it from here?

    f) cos(-2x) + 2 = 3cos(2x)

    Note: cos(-2x) = cos(2x) [proof by unit circle; or by proving cosine graph is an even function]

     cos(2x) + 2 = 3cos(2x)
     2(cos(2x)+1) = 0
    Now solve that.

    b) cos^2x + tan^2xcos^2x
    Factorise first  cos^2(x) (1+tan^2(x) )
    Use the identity  tan^2(x) + 1 = sec^2(x)
     = cos^2(x)sec^2(x)
    = 1
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  3. #3
    Newbie nadavs's Avatar
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    d) Knowing that sin^2 x + cos^2 x = 1, we can say that sin^2 x = 1 - cos^2 x. Now we can replace the 2sin^2 x with 2 - 2cos^2 x:
    2 - 2cos^2 x = 2 + cos x
    Subtract 2 from both sides:
    -2cos^2 x = cos x
    2cos^2 x + cos x = 0
    (cos^2 x)(2cos^2 x + 1) = 0
    cos^2 x = 0 or cos^2 x = -0.5
    The second case is not possible, so cos^2 x = 0, which means that:
    x = pi * k (or 180 degrees * k).

    e) Cosine is an even function, meaning that cos (-x) = cos x. That means cos (-2x) = cos (2x), so:
    cos (2x) + 2 = 3cos (2x)
    2cos (2x) = 2
    cos (2x) = 1
    2x = pi / 2 + 2pi * k (or 90 degrees + 360 degrees * k)
    x = pi / 4 + pi * k (or 45 degrees + 180 degrees * k).

    b) cos^2x + tan^2xcos^2x
    tan^2 x = (sin^2 x)/(cos^2 x). Multiply that by cos^2 x, and you'll get sin^2 x.
    As said at the beginning of the post, sin^2 x + cos^2 x = 1, so this whole expression equals 1.

    nadavs
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