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Math Help - How do I solve this equation?

  1. #1
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    How do I solve this equation?

    Could somebody help me to solve this equation. It's been a loooooong time since highschool.

    (1/sinx) + (1/cosx) = 10

    x= ??? degrees
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  2. #2
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    Re: How do I solve this equation?

    Maybe start by mulitpling through by \sin x
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  3. #3
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    Re: How do I solve this equation?

    Quote Originally Posted by taco120 View Post
    Could somebody help me to solve this equation. It's been a loooooong time since highschool.

    (1/sinx) + (1/cosx) = 10

    x= ??? degrees
    You have to solve following system of equations :

    \begin{cases}\cos x + \sin x=10\cdot \sin x\cdot \cos x \\\sin^2 x+\cos^2 x=1\end{cases}
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  4. #4
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    Re: How do I solve this equation?

    Thanks for the replies. I think I solved it. Anyone cares to doublecheck?



    \frac{1}{sin\alpha}+\frac{1}{cos\alpha}=10 \Rightarrow


    \frac{sin\alpha+cos\alpha}{sin\alpha \times cos\alpha}=10 \Rightarrow


    sin\alpha+cos\alpha=10 \times sin\alpha \times cos\alpha \Rightarrow

    Identity: sin(2\alpha)= 2 \times sin\alpha \times cos\alpha

    Times 5: 5 \times sin(2\alpha)=10 \times sin\alpha \times cos\alpha \Rightarrow

    5 \times sin(2\alpha)=sin\alpha+cos\alpha

    Squaring: 25 \times sin^2(2\alpha)=sin^2\alpha+cos^2\alpha+2 \times sin\alpha \times cos\alpha

    Identity: sin^2\alpha+cos^2\alpha=1

    \Rightarrow 25 \times sin^2(2\alpha)=1+2 \times sin\alpha \times cos\alpha

    \Rightarrow 25 \times sin^2(2\alpha)=1+sin(2\alpha) (following first identity)

    \Rightarrow 25 \times sin^2(2\alpha)-sin(2\alpha)-1=0

    x=sin(2\alpha)

    \Rightarrow 25x^2 -x -1 =0

    We now have a quadratic equation, which can be solved with the quadratic formula.
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  5. #5
    Super Member Quacky's Avatar
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    Re: How do I solve this equation?

    Looks good, the solutions will be disgusting though!
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  6. #6
    Member sbhatnagar's Avatar
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    Re: How do I solve this equation?

    Note That:There are more than 2 solutions to this equation.

    \frac{1}{\sin{x}}+\frac{1}{\cos{x}}=10

    Express the equation in terms of \cos{x}
    100\cos^4{x}-20\cos^3{x}-98\cos^2{x}+20\cos{x}-1=0

    Factorizing this (with the help of wolfram alpha):

    \begin{align*} (\cos{x}-0.993799)(\cos{x}-0.111188)(\cos{x}-0.0908748)(\cos{x}+0.995862)=1 \end{align*}

    *These are not accurate.

    We obtain \cos{x} \approx 0.993799, \hspace{4} 0.111188, \hspace{4} 0.0908748, \hspace{4} -0.995862
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