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Math Help - 2 trig problems, 0 ideas...

  1. #1
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    2 trig problems, 0 ideas...

    Find all values of x in [0,2pi]

    sin2xsinx=cosx

    and

    All values of x:

    2cos^(2) x + sinx = 1

    I'd appreciate any help! Thanks!
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  2. #2
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    Quote Originally Posted by JasonW View Post
    Find all values of x in [0,2pi]

    sin2xsinx=cosx
    For sin2xsinx = cosx

    Double Angle Identities
     2sinxcosx \times sinx - cosx = 0

    Factor
     cosx(2sin^2x - 1) = 0

    First:
     cosx = 0 (since value is 0, refer to graph)
    ANSWER:
     x = \frac {\pi}{2} and \frac {3\pi}{2}


    Second:

     2sin^2x - 1 = 0
    Simplify
     sin^2x = 1/2

     sinx = \pm 1/2

     x = sin^-1(1/2)

    Remember the CAST Rule
    ANSWER:
     x = \frac {\pi}{6} and \frac {5\pi}{6}

     x = sin^-1(-1/2)

    ANSWER:
     x = \frac {11\pi}{6} and \frac {7\pi}{6}
    Last edited by Macleef; November 24th 2007 at 09:19 PM.
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  3. #3
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    Quote Originally Posted by JasonW View Post

    and

    All values of x:

    2cos^(2) x + sinx = 1

    I'd appreciate any help! Thanks!
    Move everything to one side
    2cos^2x + sinx = 1

    Make everything in one term, only sin or only cos
    2(1 - sin^2x) + sinx - 1 = 0

    Move everything to the right side since you want your constant number for the first term to be positive
    2 - 2sin^2x + sinx - 1 = 0

    0 = 2sin^2x - sinx - 1

    This is a complex trinomial, so you need to use decomposition
    0 = 2sin^2x - 2sinx + sinx - 1

    Simplify the trinomial
    0 = 2sinx (sinx - 1) + 1 (sinx - 1)

    Set each to zero by themselves
    0 = (sinx - 1) (2sinx + 1)


    Since sinx = 1, you can find the x values by graph
    sinx - 1 =0

    sinx = 1

    ANSWER:
    x = \frac {\pi}{2}


    Determine which quadrant the angles are going to be in (CAST Rule), since the value of sin is negative, it'll be in Q3 and Q4
    2sinx + 1 =0

    sinx = \frac {-1}{2}

    x = sin^-1 \frac {-1}{2}

    Since there is a negative degrees (in negative coterminal), you need to find the related acute angle which is 30

    x = -30

    ANSWER:
    x = \frac {11\pi}{6} and \frac {7\pi}{6}
    Last edited by Macleef; November 24th 2007 at 09:27 PM.
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  4. #4
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    Thanks for the reply! A few things, though...

    In the first problem, you went from sin^2 x=1/2 to sinx=+/-1/2. Wouldn't it be +/- 1/Squareroot of 2, making the answer pi/4, not pi/6?

    Also, if I have tanx=-sinx, can I divide both sides by -sinx, and just solve what's left? (-secx=0)?

    Again, thanks for the help!
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