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Thread: another trigo equ general solution

  1. #1
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    another trigo equ general solution

    another trigo equ general solution
    p213 q11
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  2. #2
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    Quote Originally Posted by afeasfaerw23231233 View Post
    another trigo equ general solution
    p213 q11
    Your answers and the books answers are equivalent, but I much prefer your answers! Observe:

    Book answers:

    $\displaystyle \theta = \pm \frac{\pi}{2} + 2n\pi = \pm \frac{\pi}{2}, \, \frac{3\pi}{2}, \frac{5\pi}{2}, \, ....$

    $\displaystyle \theta = \frac{n \pi}{3} = 0 \, \pm \frac{\pi}{3}, \, \pm \frac{2\pi}{3}, \pm \frac{3\pi}{3} = \pm \pi, \, ....$


    These answers can be re-grouped as:

    $\displaystyle \theta = 0, \, \pm \frac{\pi}{2}, \, \pm \pi, \, \pm \frac{3\pi}{2}, \, ......$

    $\displaystyle \theta = \pm \frac{\pi}{3}, \, \frac{2 \pi}{3}, \, \frac{4 \pi}{3}, \, \frac{5 \pi}{3}, \, \frac{7 \pi}{3}, \, .....$


    The first group is given in general by $\displaystyle \theta = \frac{n \pi}{2}$ and the second group is given by $\displaystyle \theta = \pm \frac{\pi}{3} + n \pi$, where n is an integer. These are your answers.

    Which is best? It doesn't matter! (But I like yours).
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  3. #3
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    Quote Originally Posted by afeasfaerw23231233 View Post
    another trigo equ general solution
    p213 q11
    With regards to method 2, you use the fact that $\displaystyle \sin A = \sin B \Rightarrow A = B + 2n\pi$ or $\displaystyle A = (\pi - B) + 2 n \pi$.

    So $\displaystyle \sin(4 \theta) = \sin(-2 \theta)\, $ yields:

    $\displaystyle 4 \theta = -2 \theta + 2n\pi \Rightarrow 6 \theta = 2n\pi \Rightarrow \theta = \frac{n \pi}{3}$, one of the answers given in the book,

    or

    $\displaystyle 4 \theta = (\pi - [-2 \theta]) + 2n\pi = \pi + 2 \theta + 2n \pi \Rightarrow 2 \theta = \pi + 2n\pi \Rightarrow \theta = \frac{\pi}{2} + n \pi$. You can confirm that this is equivalent to $\displaystyle \theta = \pm \frac{\pi}{2} + 2n \pi$, the other answer given in the book.

    Method 2 looks like the method the book has used. But I don't know why they decided to use the general solution $\displaystyle \theta = \pm \frac{\pi}{2} + 2n \pi$ rather than $\displaystyle \theta = \frac{\pi}{2} + n \pi$ ....
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