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Math Help - trig

  1. #1
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    trig

    solve the following system of equations algebraically (-2π, 2π)

    y=cos^2 (x)
    y=sin^2 (x) + 1/2

    please show all work
    Thanks
    Last edited by meli3000; June 13th 2008 at 04:32 PM. Reason: missed part
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  2. #2
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    Cos^2(x) = Sin^2(x) + 1/2

    Cos^2(x) - Sin^2(x) = 1/2

    Cos(2x) = 1/2

    2x = Cos^-1(1/2)

    x = (Cos^-1(1/2))/2
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  3. #3
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    In fact, it's...

    pi/6, 5pi/6, -pi/6, -5pi/6
    Last edited by sean.1986; June 13th 2008 at 05:11 PM.
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  4. #4
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    Hello, meli3000!

    Solve the following system of equations algebraically (-2\pi,\:2\pi)

    . . {\color{blue}[1]}\;\;y \;=\;\cos^2\!x
    . . {\color{blue}[2]}\;\;y\:=\:\sin^2\!x + \frac{1}{2}
    \text{Add {\color{blue}[1]} and {\color{blue}[2]}: }\;2y \;=\;\underbrace{\cos^2\!x+\sin^2\!x}_{\text{This is 1}} + \frac{1}{2} \quad\Rightarrow\quad 2y \:=\:\frac{3}{2} \quad\Rightarrow\quad\boxed{y \:=\:\frac{3}{4}}

    Substitute into [1]: . \cos^2\!x \:=\:\frac{3}{4}\quad\Rightarrow\quad \cos x \:=\:\pm \frac{\sqrt{3}}{2}

    . . \boxed{x \;=\;\pm\frac{\pi}{6},\:\pm\frac{5\pi}{6},\:\pm\fr  ac{7\pi}{6},\:\pm\frac{11\pi}{6}}

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  5. #5
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    Ah yeah forgot that because it's double you get 2 more answers. Oops!
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