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Thread: 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

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

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

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

    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 $\displaystyle f(x)$ $\displaystyle n$ units to the right, you generate a function $\displaystyle g(x)=f(x-n)$. 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
    $\displaystyle 2(\sin x\cos x -\tfrac12)=0$

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

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

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

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

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

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

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

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

    Now replace $\displaystyle \theta$ by $\displaystyle 2x$, and we get

    $\displaystyle \sin 2x = 1$

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

    Divide by 2:

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

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

    Grandad
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