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Thread: Equation solutions

  1. #1
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    Equation solutions

    I want to find x-axis intercepts of this equation:
    (sin x+0.5)(sin x-0.25)=0.25

    0 <= x <= 360.

    So sin x=-0.25 or sin x=0.5

    One of the associated acute angles is 30 which is found from sin x=0.5. The other is 48.59 but I don't know where it comes from.
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  2. #2
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    hi
    put $\displaystyle \sin x=X$
    and solve $\displaystyle (X+0.5)(X-0.25)-0.25=0$
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  3. #3
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    I rearranged the equation to equal 0 using +.75 and -.5. I then solved the problem but I don't know why I couldn't solve it with the other equation.
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  4. #4
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    I think it must be necessary to factorise without any added on value. Completing the square for instance wouldn't work.
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  5. #5
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    Quote Originally Posted by Stuck Man View Post
    I think it must be necessary to factorise without any added on value. Completing the square for instance wouldn't work.

    to find the x-intercepts you must solve the given equation,

    $\displaystyle (X+0.5)(X-0.25)-0.25=X^2-0.25X+0.5X-0.375=X^2+0.25X-0.375=0$.
    i found the solutions $\displaystyle 0.5$ and $\displaystyle -0.75$
    therefore,
    $\displaystyle \sin x=0.50 \Leftrightarrow x=\arcsin (0.50)$
    or
    $\displaystyle \sin x=-0.75\Leftrightarrow x=\arcsin (-0.75)$.
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  6. #6
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    I think you've mixed up the signs of the solutions. You've rewritten the formula the same as I did. I'm still not sure why I have to fully factorise and not use the Completing The square method. I think it have something to do with the use of sin.
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  7. #7
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    Quote Originally Posted by Stuck Man View Post
    I want to find x-axis intercepts of this equation:
    (sin x+0.5)(sin x-0.25)=0.25

    0 <= x <= 360.

    So sin x=-0.25 or sin x=0.5

    One of the associated acute angles is 30 which is found from sin x=0.5. The other is 48.59 but I don't know where it comes from.
    $\displaystyle \sin^2{x} + .25\sin{x} - 0.125 = 0.25
    $

    $\displaystyle \sin^2{x} + .25\sin{x} - 0.375 = 0$

    $\displaystyle 8\sin^2{x} + 2\sin{x} - 3 = 0$

    $\displaystyle (4\sin{x} + 3)(2\sin{x} - 1) = 0$


    $\displaystyle \sin{x} = \frac{3}{4}$ ...

    $\displaystyle x = \arcsin\left(\frac{3}{4}\right) \approx 48.59^\circ$

    $\displaystyle x = 180 - \arcsin\left(\frac{3}{4}\right) \approx 131.4^\circ$


    $\displaystyle \sin{x} = \frac{1}{2}$

    $\displaystyle x = 30^\circ $

    $\displaystyle x = 150^\circ$
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  8. #8
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    I still don't know why I can't use the equation as it was originally:
    (sin x+0.5)(sin x-0.25)=0.25

    sin x=-0.25 or sin x=0.5
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  9. #9
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    Quote Originally Posted by Stuck Man View Post
    I still don't know why I can't use the equation as it was originally:
    (sin x+0.5)(sin x-0.25)=0.25

    sin x=-0.25 or sin x=0.5
    You can't do that in math.

    Now if the equation was

    $\displaystyle (\sin{x}+.5)(\sin{x}-.25) = 0$

    then you could do that

    Notice how Raoh and Skeeter set the equation equal to 0 and then solved.
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