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Trig - 2nd Degree Equations

  1. Nov 10, 2006 #1
    How do you algebraicly find the solution to this equation

    6sin(to power of 2)x-6sinx+1=0

    where 0<(or equal to) x <(or equal to) 2pi
  2. jcsd
  3. Nov 10, 2006 #2


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    Let Sinx = u and then use the binomial theorem.
  4. Nov 11, 2006 #3
    If you replace sin(x) by a variable y your equation will be :

    [tex]6y^2 -6y + 1 = 0[/tex]

    Solve this equation for y. You will get maximum two solutions for y. each solution is equal to sin(x), so you will get two goniometric equations that will be easy to solve.

  5. Nov 12, 2006 #4
    Im in trig as well and im confused with some of yalls answers. Couldn't he treat this as a quadratic function. If it were me I would see if I could factor it then set each to 0 and solve; if it doesnt factor I would plug it in the quadratic formula and solve. Im in this same chapter of trig so please let me konw what im overlooking.
  6. Nov 12, 2006 #5
    yes you can treat it as a quadratic function, thats what marlon did in his post, although you could just leave sin (x) instead of replacing it with y, I believe that just plugging sin(x) = quadratic formula would work just as well, you would then have to do inverse sin on both sides to solve for x
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