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Thread: should this factorise?

  1. #1
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    should this factorise?



    Start of problem at the top
    I believe i have done it right but it doesnt seem to factorise
    Do i need to run it from the quadratic equation?
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  2. #2
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    Quote Originally Posted by 200001 View Post


    Start of problem at the top
    I believe i have done it right but it doesnt seem to factorise
    Do i need to run it from the quadratic equation?
    $\displaystyle \sin{\theta} + \cos^2{\theta} = \frac{1}{5}$

    $\displaystyle \sin{\theta} + 1 - \sin^2{\theta} = \frac{1}{5}$

    $\displaystyle 0 = \sin^2{\theta} - \sin{\theta} - \frac{4}{5}$

    Now Complete the Square:

    $\displaystyle 0 = \sin^2{\theta} - \sin{\theta} + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 - \frac{4}{5}$

    $\displaystyle 0 = \left(\sin{\theta} - \frac{1}{2}\right)^2 - \frac{1}{4} - \frac{4}{5}$

    $\displaystyle 0 = \left(\sin{\theta} - \frac{1}{2}\right)^2 - \frac{21}{20}$

    $\displaystyle \frac{21}{20} = \left(\sin{\theta} - \frac{1}{2}\right)^2$

    $\displaystyle \pm\sqrt{\frac{21}{20}} = \sin{\theta} - \frac{1}{2}$

    $\displaystyle \pm \frac{\sqrt{21}}{2\sqrt{5}} = \sin{\theta} - \frac{1}{2}$

    $\displaystyle \pm \frac{\sqrt{105}}{10} = \sin{\theta} - \frac{1}{2}$

    $\displaystyle \frac{1}{2} \pm \frac{\sqrt{105}}{10} = \sin{\theta}$

    $\displaystyle \frac{5 \pm \sqrt{105}}{10} = \sin{\theta}$

    $\displaystyle \theta = \arcsin{\left(\frac{5 + \sqrt{105}}{10}\right)}$ or $\displaystyle \theta = \arcsin{\left(\frac{5 - \sqrt{105}}{10}\right)}$.
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  3. #3
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    Yes, thats where i went with it but this is from a book that isnt driven at any equations that wont factorise and the level its picthed at would not go into that depth.
    I had the same answer yet though I had gone wrong somewhere
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  4. #4
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    Hello 200001
    Quote Originally Posted by 200001 View Post


    Start of problem at the top
    I believe i have done it right but it doesnt seem to factorise
    Do i need to run it from the quadratic equation?
    You have a sign wrong in the third line of your solution. It should be:
    $\displaystyle \sin^2\phi-\sin\phi=\frac45$

    $\displaystyle \Rightarrow 5\sin^2\phi-5\sin\phi-4=0$

    $\displaystyle \Rightarrow \sin\phi = \frac{5\pm\sqrt{105}}{10}$
    Now $\displaystyle -1\le\sin\phi\le 1$. So the only valid root is:
    $\displaystyle \sin\phi = \frac{5-\sqrt{105}}{10}$

    $\displaystyle \Rightarrow \phi = n\pi +(-1)^n \arcsin\left(\frac{5-\sqrt{105}}{10}\right)$
    Grandad
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  5. #5
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    Yes
    that makes sense
    Thanks!
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