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Thread: it should be easy by now

  1. #1
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    it should be easy by now

    Hi folks,

    you can see that I have posted quite a few trig questions recently, so I should have it by now!

    But sadly no!

    say I have cos 2x = 0 and I want the general solution.

    $2x = cos^{-1}(0) = 90$ and 270 degrees

    so $2x = \dfrac{\pi}{2} + n\pi$ where $n \in \mathcal{Z}$

    therefore $x = \dfrac{\pi}{4} + \dfrac{n\pi}{2}$

    but this is wrong!

    The answer is $n\pi+ \dfrac{\pi}{4}$

    in my answer, I put the general term in before dividing by 2 and this seems the way you guys have been doing it in previous posts.
    So what's wrong this time?
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  2. #2
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    Re: it should be easy by now

    I agree with your solution, not "theirs" ...

    the cosine function equals zero at odd-multiples of $\dfrac{\pi}{2}$

    $\cos(2x)=0 \implies 2x=\dfrac{\pi}{2}(2n+1), \, n\in \mathbb{Z}$

    $x=\dfrac{\pi}{4}(2n+1) = \dfrac{\pi}{4}+\dfrac{n\pi}{2}$

    so, $x$ is all odd-multiples of $\dfrac{\pi}{4}$

    consider $x=\dfrac{3\pi}{4} \implies \cos(2x) = \cos\left(\dfrac{3\pi}{2}\right) =0$

    "their" solution skips that value (and other ones) for $x$ ...

    $\dfrac{\pi}{4} + n\pi = \dfrac{\pi}{4}, \dfrac{5\pi}{4}, \dfrac{9\pi}{4}, \, ...$
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  3. #3
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    Re: it should be easy by now

    Your answer is correct. The answer given is only half right, it would need to be $\displaystyle \begin{align*} \left\{ \frac{\pi}{4} , \frac{3\,\pi}{4} \right\} + n\,\pi \end{align*}$.
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  4. #4
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    Re: it should be easy by now

    Thanks to skeeter (again) and to Prove It.

    I was on the brink of giving up maths altogether!

    You have restored my faith.

    Thanks Gents
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  5. #5
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    Re: it should be easy by now

    If you are planning on taking calculus, I encourage you to work overtime and become comfortable with trig. You will encounter a lot of trigonometry in calculus.
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  6. #6
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    Re: it should be easy by now

    Thanks for the encouragement.
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