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Math Help - please solve the trig equation

  1. #1
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    please solve the trig equation

    sin2xcosx+cos2xsinx=1
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  2. #2
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    Use the double angle identities, I have used the sin form of cos(2x) because I want to eliminate cos

    2\sin(x)\cos(x)\cos(x) + (1- 2\sin^2(x))\sin(x) = 1 \implies 2\sin(x) \cos^2(x) + \sin(x) - 2\sin^3(x) = 1

    Of course \cos^2(x) = 1-\sin^2(x) and hence we have

    2\sin(x) (1-\sin^2(x)) + \sin(x) - 2\sin^3(x) = 1 \implies 2\sin(x) - 2\sin^3(x) + \sin(x) - 2\sin^3(x) = 1
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    \displaystyle \sin 2x \cos x+\cos 2x \sin x = \sin (2x+x)
    Last edited by pickslides; February 16th 2011 at 12:33 PM. Reason: bad latex...
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    Use the double angle identities, I have used the sin form of cos(2x) because I want to eliminate cos

    2\sin(x)\cos(x)\cos(x) + (1- 2\sin^2(x))\sin(x) = 1 \implies 2\sin(x) \cos^2(x) + \sin(x) - 2\sin^3(x) = 1

    Of course \cos^2(x) = 1-\sin^2(x) and hence we have

    2\sin(x) (1-\sin^2(x)) + \sin(x) - 2\sin^3(x) = 1 \implies 2\sin(x) - 2\sin^3(x) + \sin(x) - 2\sin^3(x) = 1
    I followed your math up to where you stopped. What do you do now? At first I thought the two cubed sins will cancel each other out but they're both -2sin^3, so what do you do next? please respond. thanks!!!
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  5. #5
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    Quote Originally Posted by TacticalPro View Post
    I followed your math up to where you stopped. What do you do now? At first I thought the two cubed sins will cancel each other out but they're both -2sin^3, so what do you do next? please respond. thanks!!!
    Rewrite the equation as \displaystyle 2X - 2X^3 + X - 2X^3 = 1, with \displaystyle X = \sin{x}.

    Solve for \displaystyle X, then use this to solve for \displaystyle {x}.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    Rewrite the equation as \displaystyle 2X - 2X^3 + X - 2X^3 = 1, with \displaystyle X = \sin{x}.

    Solve for \displaystyle X, then use this to solve for \displaystyle {x}.
    Ok i replaced sinx with x and simplified it to (-4x^3) + (3x) - 1

    what do I do next? I tried factoring out an x but got no where
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  7. #7
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    Use the Factor Theorem to find a linear factor and use long division to factorise this cubic to its linear and quadratic factors. Factorise the quadratic if possible. Then solve using Null Factor Law.
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  8. #8
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    Quote Originally Posted by Prove It View Post
    Use the Factor Theorem to find a linear factor and use long division to factorise this cubic to its linear and quadratic factors. Factorise the quadratic if possible. Then solve using Null Factor Law.
    I'm sorry but I have no idea what you just said. I looked up factor theorem and had no luck deciphering. could you please explain further? thanks
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  9. #9
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    Quote Originally Posted by TacticalPro View Post
    I'm sorry but I have no idea what you just said. I looked up factor theorem and had no luck deciphering. could you please explain further? thanks
    Post #2 tells you exactly what to do. Why would you want to do it in a much more complicated way. When you want to kill a fly, use a fly swatter. Not an elephant gun.
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    Post #2 tells you exactly what to do. Why would you want to do it in a much more complicated way. When you want to kill a fly, use a fly swatter. Not an elephant gun.
    I looked at post 2 but don't understand it? all he did was change the 1 to a sin(2x+x)?? what identity supports that? what is he implying?
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  11. #11
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    Quote Originally Posted by TacticalPro View Post
    I looked at post 2 but don't understand it? all he did was change the 1 to a sin(2x+x)?? what identity supports that? what is he implying?
    The identity they used is \displaystyle \sin{(A + B)} \equiv \sin{A}\cos{B} + \cos{A}\sin{B}.

    Here \displaystyle A = 2x and \displaystyle B = x.
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  12. #12
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    Quote Originally Posted by Prove It View Post
    The identity they used is \displaystyle \sin{(A + B)} \equiv \sin{A}\cos{B} + \cos{A}\sin{B}.

    Here \displaystyle A = 2x and \displaystyle B = x.
    Ok then, i use that to get

    sin(2x+x) = 1

    what can i do from that?
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  13. #13
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    Surely you can solve \displaystyle \sin{3x} = 1...
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  14. #14
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    Quote Originally Posted by Prove It View Post
    Surely you can solve \displaystyle \sin{3x} = 1...
    shoot me.... i cramped, thought it was sin^(2x+x)=1

    so anyway it should be:

    sin2x+sinx=1

    sin3x=1

    sinx=1/3
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  15. #15
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    Quote Originally Posted by TacticalPro View Post
    shoot me.... i cramped, thought it was sin^(2x+x)=1

    so anyway it should be:

    sin2x+sinx=1

    sin3x=1

    sinx=1/3
    Absolutely not!

    \displaystyle \sin{3x} = 1 \implies 3x = \arcsin{1} + 2\pi n with \displaystyle n \in \mathbf{Z}...
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