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Math Help - Easy trig question

  1. #1
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    Easy trig question

    It's been awhile since I've done high school math, help would be greatly appreciated!

    Solve the following for x:

    2sin^2(x)=1

    0 \leq x <2\pi

    The solution set is... (should be 4)
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  2. #2
    Master Of Puppets
    pickslides's Avatar
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    Looks better like this

    2\sin^2(x)=1

    \sin^2(x)=\frac{1}{2}

    \sin(x)=\pm\frac{1}{\sqrt{2}}

    Spoiler:
    x = \frac{\pi}{4} and?
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  3. #3
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    Wouldn't it be that

    sin(x) = \pm \frac{\sqrt{2}}{2}?

    I think I got it now though.
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  4. #4
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    \displaystyle \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}
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