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Thread: Two sin equations

  1. #1
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    Two sinus equations

    The first equation is

    $\displaystyle 75*\sin{x}+5=0$

    I tried solving it:
    $\displaystyle \sin{x}=\frac{-5}{75}$
    $\displaystyle x1 = -3.82^{\circ} + 360K$
    $\displaystyle x2 = 183.82^{\circ} + 360K$

    But the book says it's
    $\displaystyle x1 = -45.58^{\circ} + 360K$
    $\displaystyle x2 = 225.58^{\circ} + 360K$


    The second equation looks more simple and I can solve it with my calculator but I want to know how to solve it without my calculator

    $\displaystyle \sin{3x} = 1$


    Thank you
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  2. #2
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    You have gotten to $\displaystyle \displaystyle \sin{x} = -\frac{1}{15}$. You need to remember that sine is negative in the third and fourth quadrants.

    So your solutions are $\displaystyle \displaystyle \left\{ 180^{\circ} + \sin^{-1}{\left(\frac{1}{15}\right)}, 360^{\circ} - \sin^{-1}{\left(\frac{1}{15}\right)}\right\} + 360^{\circ}K$.


    For the second, $\displaystyle \displaystyle \sin{X} = 1$ if $\displaystyle \displaystyle X = 90^{\circ} + 360^{\circ}K$.

    So if $\displaystyle \displaystyle \sin{3x} = 1$

    $\displaystyle \displaystyle 3x = 90^{\circ} + 360^{\circ}K$

    $\displaystyle \displaystyle x = 30^{\circ} + 120^{\circ}K$.
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  3. #3
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    Quote Originally Posted by klik11 View Post
    The first equation is

    $\displaystyle 75*\sin{x}+5=0$

    I tried solving it:

    $\displaystyle \sin{x}=\frac{-5}{75}$

    $\displaystyle x1 = -3.82^{\circ} + 360K$

    $\displaystyle x2 = 183.82^{\circ} + 360K$

    But the book says it's

    $\displaystyle x1 = -45.58^{\circ} + 360K$

    $\displaystyle x2 = 225.58^{\circ} + 360K$


    Thank you
    The book answers are the solution to

    $\displaystyle 7sinx+5=0$
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