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Math Help - √3.cosX sinX = √3, How To Solve This?

  1. #1
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    Question √3.cosX sinX = √3, How To Solve This?

    could you help me please... √3.cosX sinX = √3
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  2. #2
    Junior Member bebrave's Avatar
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    firstly divide all sgrt3
    then when you see sgrt 3 you will be write tanpi/3 then you can solve...or if you want i can solve all
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  3. #3
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    can you please do it ?
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    Junior Member bebrave's Avatar
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    ok..wait me pls 10 minute..
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  5. #5
    Junior Member bebrave's Avatar
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    i have like this question and solution you can check attachment then you can solve your problem
    Attached Thumbnails Attached Thumbnails √3.cosX  sinX = √3, How To Solve This?-trigo.jpg  
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  6. #6
    Moo
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    P(I'm here)=1/3, P(I'm there)=t+1/3
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    Hi,

    There's a simpler way.

    When you don't know where to start, when given a\cos(x)+b\sin(x), find the modulus, that is \sqrt{a^2+b^2} and factor it out.
    You would then have \sqrt{a^2+b^2} \left(\frac{a}{\sqrt{a^2+b^2}} \cos(x)+\frac{b}{\sqrt{a^2+b^2}} \sin(x)\right)

    Define c=\frac{a}{\sqrt{a^2+b^2}} and d=\frac{b}{\sqrt{a^2+b^2}}
    they're both in [-1,1] and are such that c^2+d^2=1
    Thus there exists \theta \in [0,2\pi[ such that c=\cos \theta and d=\sin \theta

    Hence a\cos x+b\sin x=\cos\theta\cos x+\sin\theta\sin x=\cos(x-\theta)


    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    Here, you can find that \sqrt{a^2+b^2}=2

    So you can divide by 2 on both sides :
    \frac{\sqrt{3}}{2}\cos(x)-\frac 12 \sin(x)=\frac{\sqrt{3}}{2}

    Note that \cos\left(-\tfrac\pi 6\right)=\tfrac{\sqrt{3}}{2} and \sin\left(-\tfrac\pi 6\right)=-\tfrac 12

    So your equation is now \cos\left(-\tfrac\pi 6\right)\cos(x)+\sin\left(-\tfrac\pi 6\right)\sin(x)=\tfrac{\sqrt{3}}{2}

    That is \cos\left(x+\tfrac\pi 6\right)=\tfrac{\sqrt{3}}{2}



    Then remember that \cos(a)=\cos(b) \Leftrightarrow \forall k\in \mathbb{Z} ~,~ a=\pm b+2k\pi
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