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Math Help - [SOLVED] trig function cos^2

  1. #1
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    [SOLVED] trig function cos^2

    Solve: 4 cos^2 x = 3
    for
    0 < x < 2pi

    I thought the way I did it was right but I get the wrong answer.

    so I factored
    cosx(4cosx)=3
    so
    cosx=3 and 4cosx=3
    cosx cannot =3 so 4cosx=3 ....
    cosx=3/4

    so I sketch a little graph and find x=pi/6 and 15 pi/8? but this gets confusing. whats the right way to find this?
    but that isnt right, could someone help me with this question?

    thanks.
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  2. #2
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    Quote Originally Posted by brentwoodbc View Post
    Solve: 4 cos^2 x = 3
    for 0 < x < 2pi
    No not quite.
    4 cos^2(x)=3
    cos(x)=\pm \frac{\sqrt{3}}{2}
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  3. #3
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    Quote Originally Posted by Plato View Post
    No not quite.
    4 cos^2(x)=3
    cos(x)=\pm \frac{\sqrt{3}}{2}
    thanks I just realized that.
    but my teacher said you could factor and solve like.
    cos^2x-1=0
    cos(x)=0 and cosx=1
    x=0,pi/2,3pi/2,2pi ? is that wrong?
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  4. #4
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    Quote Originally Posted by brentwoodbc View Post
    my teacher said you could factor and solve like. cos^2x-1=0
    cos(x)=0 and cosx=1
    x=0,pi/2,3pi/2,2pi ? is that wrong?
    Yes it is wrong.
    4cos^2(x)-3=0
    cos^2(x)-\frac{3}{4}=0
    (cos(x)-\frac{\sqrt3}{2})(cos(x)+\frac{\sqrt3}{2})=0
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  5. #5
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    sorry an example of the way I did it is
    sin^2x-sinx=0
    sinx = 0 and sinx-2=0

    this is one of the ways he taught us, he said if its a quadratic to do this, isnt the first example I gave you a quadratic with the b term =0 so why doesnt it work?
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  6. #6
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    Quote Originally Posted by Plato View Post
    Yes it is wrong.
    4cos^2(x)-3=0
    cos^2(x)-\frac{3}{4}=0
    (cos(x)-\frac{\sqrt3}{2})(cos(x)+\frac{\sqrt3}{2})=0
    thanks that answers my last post.

    Thank you.
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