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Math Help - Easy Trigonometry (5 questions)

  1. #1
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    Exclamation Easy Trigonometry (5 questions)

    Solve each equation on the interval [0,2pi]

    1) sin^2x-2cosx-2=0

    2) 4cos^x=5-4sinx

    3) sin2x=cosx

    4) cos2x=cosx

    5) cos2x+5cosx+3=0
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  2. #2
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    Re: Easy Trigonometry (5 questions)

    Quote Originally Posted by LillyJackson06 View Post
    Solve each equation on the interval [0,2pi]

    1) sin^2x-2cosx-2=0

    2) 4cos^x=5-4sinx

    3) sin2x=cosx

    4) cos2x=cosx

    5) cos2x+5cosx+3=0
    Please show what work you have been able to do. We are not a homework service.

    -Dan
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  3. #3
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    Re: Easy Trigonometry (5 questions)

    I do not know how to start these equations, and what to substitute in, which is the first step, which i need help on.
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    Re: Easy Trigonometry (5 questions)

    Quote Originally Posted by LillyJackson06 View Post
    Solve each equation on the interval [0,2pi]
    1) sin^2x-2cosx-2=0
    2) 4cos^x=5-4sinx
    3) sin2x=cosx
    4) cos2x=cosx
    5) cos2x+5cosx+3=0

    Write #1 as 1-\cos^2(x)-2\cos(x)-2=\cos^2(x)+2\cos(x)+1=0.
    Then let u=\cos(x) solve u^2+2u+1=0.

    You have left something out of #2.

    #3 \sin(2x)=2\sin(x)\cos(x)

    #5 \cos(2x)=2\cos^2(x)-1
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  5. #5
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    Re: Easy Trigonometry (5 questions)

    I now understand #1.

    Sorry, #2 should be 4cos^2x=5-4sinx.

    I also now get #3 and #5.

    Thanks so much.
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  6. #6
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    Re: Easy Trigonometry (5 questions)

    Quote Originally Posted by LillyJackson06 View Post
    I now understand #1.

    Sorry, #2 should be 4cos^2x=5-4sinx.

    I also now get #3 and #5.

    #2 4(1-\sin^2(x))=5-4\sin(x)
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