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Math Help - general solution?

  1. #1
    Newbie Viky's Avatar
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    Post general solution?

    determine the general solution:

    2(sinxcosx - 1/2) =0


    2} Write down the equation of q, if q is the result of p shifted 3 units to the right.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Viky View Post
    determine the general solution:

    2(sinxcosx - 1/2) =0
    Note that when you distribute the 2, you have

    2\sin x\cos x-1=0\implies \sin(2x)=1

    Thus, 2x=\sin^{-1}(1)\implies 2x=\frac{(4k-3)\pi}{2};\,k\in\mathbb{Z}

    So it follows that x=\frac{(4k-3)\pi}{4};\,k\in\mathbb{Z}

    Does this make sense?

    2} Write down the equation of q, if q is the result of p shifted 3 units to the right.
    When you shift a function f(x) n units to the right, you generate a function g(x)=f(x-n). Use this idea to answer this part of the question.
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  3. #3
    Newbie Viky's Avatar
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    i didn't get the first part
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  4. #4
    MHF Contributor
    Grandad's Avatar
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    General solution of trig equation

    Hello Viky
    Quote Originally Posted by Viky View Post
    i didn't get the first part
    2(\sin x\cos x -\tfrac12)=0

    \Rightarrow 2\sin x \cos x -1 =0

    \Rightarrow \sin 2x = 1, using the identity \sin 2x = 2 \sin x \cos x

    Since \sin \frac{\pi}{2} = 1, one possible solution is 2x = \frac{\pi}{2}; i.e. x = \frac{\pi}{4}. But how do we find the general solution? Like this:

    The positive values of \theta that make \sin\theta = 1 are \frac{\pi}{2},\frac{\pi}{2}+ 2\pi,\frac{\pi}{2}+ 4\pi, \frac{\pi}{2}+ 6\pi, and so on.

    Or we can go the other way and find negative values of \theta by taking away multiples of 2\pi. So \sin\theta = 1 is also satisfied by \theta = \frac{\pi}{2}- 2\pi, \frac{\pi}{2}-4\pi, and so on.

    Combining all these together we can say that the general solution of \sin\theta = 1 is:

    \theta = \frac{\pi}{2}+ 2k\pi, k = 0, \pm1, \pm2, \pm3, ...

    or simply \theta = \frac{\pi}{2}+ 2k\pi, k \in \mathbb{Z}, where \mathbb{Z} = \{\text{integers}\}

    Now replace \theta by 2x, and we get

    \sin 2x = 1

    \Rightarrow 2x = \frac{\pi}{2}+ 2k\pi, k \in \mathbb{Z}

    Divide by 2:

    \Rightarrow x = \frac{\pi}{4}+ k\pi, k \in \mathbb{Z}

    This is equivalent to the answer that Chris L T521 gave you, but the values of k start in a different place.

    Grandad
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