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Math Help - finding x value in trig equation

  1. #1
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    finding x value in trig equation

    hello can someone please demonstrate how a question like this is solved for the the value of (x) between 0 and 360 degrees.

     <br />
2-sinx=cos^2x +7sin^2x<br />
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  2. #2
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    Quote Originally Posted by sigma1 View Post
    hello can someone please demonstrate how a question like this is solved for the the value of (x) between 0 and 360 degrees.

     <br />
2-sinx=cos^2x +7sin^2x<br />
    Start by substituting \cos^2 x = 1 - \sin^2 x. Then re-arrange the resulting equation into a quadratic equation where \sin x is the variable. Solve this equation for \sin x and then solve for x.

    If you need more help, please show all your work and say where you get stuck.
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  3. #3
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    ok i have

     <br />
2-sinx =1-sin^2x =7sin^2x<br />
     <br />
2-sinx= 1-6sin^2x<br />

     <br />
-sinx=-6sin^2x<br />

    i cannot seem to get a value between 0 and 360 for my angle i suspect i did something wrong along the way... can you show me how its done..thanks..
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  4. #4
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    Quote Originally Posted by sigma1 View Post
    ok i have

     <br />
2-sinx =1-sin^2x =7sin^2x<br />
     <br />
2-sinx= 1-6sin^2x<br />
    What happen to your constant?
    2-sin(x)=1-6sin^{2}(x)\rightarrow-sin(x)=-1-6sin^{2}(x)
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