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    Trigonometry

    Find all the angles, in terms of pi, for 0<x<5 which satisfy the equation
    sin2x=sin(x-4)
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  2. #2
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    Quote Originally Posted by acc100jt View Post
    Find all the angles, in terms of pi, for 0<x<5 which satisfy the equation
    sin2x=sin(x-4)
    If $\displaystyle \sin{2x} = \sin{(x - 4)}$ then $\displaystyle 2x = x - 4$ in the domain $\displaystyle 0<x<5$.

    Surely you can solve that...
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  3. #3
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    Trig equations

    Hello acc100jt
    Quote Originally Posted by acc100jt View Post
    Find all the angles, in terms of pi, for 0<x<5 which satisfy the equation
    sin2x=sin(x-4)
    In general, if $\displaystyle \sin A = \sin B$, then $\displaystyle A = n\pi + (-1)^nB$

    For example, when $\displaystyle n=0, A =B$

    When $\displaystyle n = 1, A = \pi - B$

    When $\displaystyle n = 2, A = 2\pi + B$

    When $\displaystyle n = -1, A = -\pi - B$

    ... and so on.

    So, what you have to do is to solve the equation $\displaystyle 2x = n\pi +(-1)^n(x-4) $ for all the values of $\displaystyle n$ that give values of $\displaystyle x$ in the required range.

    So try the values of $\displaystyle n$ that I've given above: $\displaystyle 0, 1, 2, -1,$ ...etc and see what happens.

    Grandad
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