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Math Help - 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:

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

    \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:

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

    \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 \theta = \frac{n \pi}{2} and the second group is given by \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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    Quote Originally Posted by afeasfaerw23231233 View Post
    another trigo equ general solution
    p213 q11
    With regards to method 2, you use the fact that \sin A = \sin B \Rightarrow A = B + 2n\pi or A = (\pi - B) + 2 n \pi.

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

    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

    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 \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 \theta = \pm \frac{\pi}{2} + 2n \pi rather than \theta = \frac{\pi}{2} + n \pi ....
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