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Math Help - solving trig equations

  1. #1
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    solving trig equations

    Can somebody explain to me how to solve this please?
    tan2x=(cos2x)/2
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by unmcintosh View Post
    Can somebody explain to me how to solve this please?
    tan2x=(cos2x)/2
    \tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{\cos(2x)}  {2}

    cross multipliing we get

    2\sin(2x)=\cos^2(2x)=1-\sin^2(2x)\Rightarrow{\sin^2(2x)+2\sin2x)-1=0}

    Now let \sin(2x)=u

    giving us

    u^2+2u-1

    solving by quadratic equation we get

    \frac{-2\pm\sqrt{4-4(1)(1)}}{2}

    can you go from there?
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  3. #3
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    yes i can,
    thank you!
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by unmcintosh View Post
    yes i can,
    thank you!
    Note, that once you get the answer for the quadratic, you must do the following

    Let the solution of the qudratic be u=\pm{c}

    since you want the answer for x, you go back to yoru sub

    u=\sin(2x)=\pm{c}\Rightarrow{x=\frac{\arcsin\pm{c}  }{2}}
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    \tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{\cos(2x)}  {2}

    cross multipliing we get

    2\sin(2x)=\cos^2(2x)=1-\sin^2(2x)\Rightarrow{\sin^2(2x)+2\sin2x)-1=0}

    Now let \sin(2x)=u

    giving us

    u^2+2u-1

    solving by quadratic equation we get

    \color{red}\boxed{\frac{-2\pm\sqrt{4-4(1)(1)}}{2}}

    can you go from there?
    Inside the square root, it should be \sqrt{4-4(1)(-1)}.

    The way you did it would give one solution, when in fact there are two solutions.
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