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Math Help - Trig Function Intersection

  1. #1
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    Trig Function Intersection

    Give the exact intersection point for the equations f(x)=4sin^2x+7sinx+6 and g(x)=2cos^2x-4sinx+11

    Ok, my result is that there is no intersection point because if you put f(x)=g(x) and try to solve for x or the intersection point, the LS f(x) is not possible, so there is none?


    OK can someone just clarify for me if I have this right? Explain please thanks!
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  2. #2
    Senior Member Twig's Avatar
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    hi

    hi

    4 sin^{2}(x)+7 sin(x)+6=2 cos^{2}(x)-4sin(x)+11

    Let  u = sin(x) \mbox{ then we get because } cos^{2}(x) = 1-sin^{2}(x) \, \, \; \; 6u^{2}+11u-7=0

    This has the only valid solution  u = \frac{1}{2}
    But  u=sin(x)=\frac{1}{2} \Rightarrow x = \frac{\pi}{6} is a solution. Are there more?
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  3. #3
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    Quote Originally Posted by Twig View Post
    hi

    4 sin^{2}(x)+7 sin(x)+6=2 cos^{2}(x)-4sin(x)+11

    Let  u = sin(x) \mbox{ then we get because } cos^{2}(x) = 1-sin^{2}(x) \, \, \; \; 6u^{2}+11u-7=0

    This has the only valid solution  u = \frac{1}{2}
    But  u=sin(x)=\frac{1}{2} \Rightarrow x = \frac{\pi}{6} is a solution. Are there more?
    How did you get from 2(1-sin^2x)-4sinx+11 to 6cos^2x+11cosx-7?????
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  4. #4
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    Quote Originally Posted by skeske1234 View Post
    How did you get from 2(1-sin^2x)-4sinx+11 to 6cos^2x+11cosx-7?????
    No

     u = \sin(x)

    4 \sin^{2}(x)+7 \sin(x)+6=2 (1-\sin^{2}(x))-4\sin(x)+11

    4 u^{2}+7 u+6=2 (1-u^{2})-4u+11

    6u^{2}+11u-7=0
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  5. #5
    Senior Member Twig's Avatar
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    Maybe I should have been more clear, my apologies.
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