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Math Help - solving sine

  1. #1
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    solving sine

    i've been having problems with these...

    sin x= -.5 ,0 < x <2pi

    sinx =-1/2

    x=11pi/6 +2kpi
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  2. #2
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    Quote Originally Posted by cumana View Post
    i've been having problems with these...

    sin x= -.5 ,0 < x <2pi

    sinx =-1/2

    x=11pi/6 +2kpi
    Not quite. Try thinking of it this way: We get a reference angle of x = \frac{\pi}{6}.

    Now, sine is negative in two quadrants, III and IV. So we have solutions for sin(x) = -\frac{1}{2}:
    x = \frac{7 \pi}{6}
    and
    x = \frac{11 \pi}{6}

    So the solution is
    x = \frac{7 \pi}{6} + 2k \pi
    and
    x = \frac{11 \pi}{6} + 2m \pi
    (You really don't need to change the arbitrary integer from k to m, but I do it to remind myself that the two do not need to be the same.)

    -Dan
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  3. #3
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    Hello, cumana!

    \sin x\:=\:-\frac{1}{2},\;\;{\bf{\color{red}0< x < 2\pi}}

    Since \sin\frac{\pi}{6} \:=\:\frac{1}{2}, and sine is negative in Quadrants 3 and 4,

    . . then: . \boxed{x \;=\;\frac{7\pi}{6},\:\frac{11\pi}{6}}


    Those are the only answers in the given interval.

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