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Math Help - sin(2x) minus square root of three over two

  1. #1
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    Post sin(2x) minus square root of three over two

    Sin(2x) - √3/2 **Read as sin(2x) minus square root of three over two** Instructions say to find ALL solutions. However, I believe I am coming up with some negative erroneous solutions.


    (2sinx cosx)^2 = (√3/2)^2 **Square both sides**
    4sin^2x cos^2x =
    4sin^2x (1 – sin^2x) = **substitute cos2x = 1 – sin2x**
    4sin^2x – 4sin^4x = **distribute**
    -4sin^4 + 4sin^2x = ** re-arrange by descending powers**
    -4sin^4 + 4sin^2x – = 0 **now quadratic in form.**
    -4u^2 + 4u – =0 **let “u” equal sin2**
    U = and
    Sin^2x = and sin^2x = **take the square root of both sides**

    Then sinx = plus or minus √3/2 and sinx = plus or minus √1/2

    I believe the quadratic is confusing me due to the plus or minus issue. So I am looking for a way that might be simpler, and quicker without the quadratic.






    I have another problem that is very similar:


    sin(2x) = 1/2

    Thanks in advance.
    Last edited by lawnman2006; October 21st 2010 at 11:08 PM. Reason: power raised in the wrong place
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  2. #2
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    \displaystyle \sin{2x} = \frac{\sqrt{3}}{2}

    \displaystyle 2x = \arcsin{\frac{\sqrt{3}}{2}}

    \displaystyle 2x = \left\{ \frac{\pi}{3}, \frac{2\pi}{3}\right\} + 2\pi n where n \in \mathbf{Z}

    \displaystyle x = \left\{\frac{\pi}{6}, \frac{\pi}{3}\right\} + \pi n.


    Follow the same process for your second equation.
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    Thank you!
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