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Math Help - Find exact solutions algebraically?

  1. #1
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    Find exact solutions algebraically?

    How do I find the exact solutions algebraically for the following equation?
    sin 2x sin x = cos x

    Any help is appreciated.
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  2. #2
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    Quote Originally Posted by iluvmathbutitshard View Post
    How do I find the exact solutions algebraically for the following equation?
    sin 2x sin x = cos x

    Any help is appreciated.
    sin(2x)=2sin(x)cos(x). That should simplify to something easier to solve.
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  3. #3
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    Ok, so how would I simplify further?
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  4. #4
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    Quote Originally Posted by iluvmathbutitshard View Post
    Ok, so how would I simplify further?
    Substitute for sin(2x) in your problem. cos(x) will cancel, divide by 2 and you'll get something simpler. I don't understand where you are getting stuck...
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  5. #5
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    Ok, thank you. I got it now.
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    Quote Originally Posted by Jameson View Post
    Substitute for sin(2x) in your problem. cos(x) will cancel, divide by 2 and you'll get something simpler. I don't understand where you are getting stuck...
    Without knowing the domain you could potentially be dividing by 0 when cos(x)=0 if you cancel. It is better to take cos(x) from both sides and factor out cos(x)

    2sin^2(x)cos(x)-cos(x)=0

    <br />
cos(x)(2sin^2(x)-1)=0

    Either cos(x)=0 or 2sin^2(x)-1=0

    This should give 3 principal solutions and there may be more or less depending on the domain (usually 0 \leq x \leq 2\pi )
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  7. #7
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    Very good observation. Thank you for pointing this out.
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