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Math Help - Solving Trig Identities??

  1. #1
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    Solving Trig Identities??

    For my homework tonight, there is this one problem that is supposed to be easy but i can't figure it out.

    Solve for exact solutions in the interval [0degrees, 360degrees)
    [ means include, & ) means exclude
    2sinx -1= cscx
    I am supposed to find the degree value for x. and i think they want me to use cscx=1/sinx
    i'm not sure, i already tried and it didn't work. Please help me!!
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  2. #2
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    Quote Originally Posted by shanzwhaat View Post
    For my homework tonight, there is this one problem that is supposed to be easy but i can't figure it out.

    Solve for exact solutions in the interval [0degrees, 360degrees)
    [ means include, & ) means exclude
    2sinx -1= cscx
    I am supposed to find the degree value for x. and i think they want me to use cscx=1/sinx
    i'm not sure, i already tried and it didn't work. Please help me!!
    They do! Then multiply by \sin x and you have a quadratic equation in \sin x which you can solve.
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    Thanks, but i did that and got 0degrees, 180degrees, 30 and 150
    but the back of the book said the answers were 90degrees, 210degrees, and 330degrees.. Do you know what I'm doing wrong??/
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  4. #4
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    Quote Originally Posted by shanzwhaat View Post
    Thanks, but i did that and got 0degrees, 180degrees, 30 and 150
    but the back of the book said the answers were 90degrees, 210degrees, and 330degrees.. Do you know what I'm doing wrong??/
    If 2\sin x - 1 = \csc x then 2\sin x - 1 = \frac{1}{\sin x} so

    2 \sin^2 x - \sin x -1 = 0 or (2 \sin x + 1)(\sin x -1) = 0\;\;\Rightarrow\;\; \sin x = - \frac{1}{2}, 1

    Did you get that?
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  5. #5
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    Quote Originally Posted by shanzwhaat View Post
    For my homework tonight, there is this one problem that is supposed to be easy but i can't figure it out.

    Solve for exact solutions in the interval [0degrees, 360degrees)
    [ means include, & ) means exclude
    2sinx -1= cscx
    I am supposed to find the degree value for x. and i think they want me to use cscx=1/sinx
    i'm not sure, i already tried and it didn't work. Please help me!!
    2\sin{x} - 1 = \csc{x}

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

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

    2\sin^2{x} - \sin{x} - 1 = 0.


    Let X = \sin{x} so

    2X^2 - X - 1 = 0.

    2X^2 - 2X + X - 1 = 0

    2X(X - 1) + 1(X - 1) = 0

    (X - 1)(2X + 1) = 0

    X - 1 = 0 or 2X + 1 = 0

    X = 1 or X = -\frac{1}{2}


    So \sin{x} = 1 or \sin{x} = -\frac{1}{2}.


    In the domain x \in [0^{\circ}, 360^{\circ}), \sin{x} = 1 if x = 90^{\circ}.

    In the domain x \in [0^{\circ}, 360^{\circ}), \sin{x} = -\frac{1}{2} if x = 210^{\circ} (Quadrant 3) or x = 330^{\circ} (Quadrant 4).


    Does that make sense?
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  6. #6
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    YES! thank you everyone who answered!
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