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Math Help - Solving Trig Equations

  1. #1
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    Solving Trig Equations

    Hi guys,
    I have 2 equations on my homework that I am not allowed to use a calculator on, and need to find the exact answer. Sorry im posting so much, I got a gigantic packet to work on.

    2sin^{2}(x)+5cos(x)=4where 0 \le x\le360

    AND

    sin(2x)=2cos^{2}(x)where 0\le x\le 2PI
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  2. #2
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    Quote Originally Posted by spycrab View Post
    Hi guys,
    I have 2 equations on my homework that I am not allowed to use a calculator on, and need to find the exact answer. Sorry im posting so much, I got a gigantic packet to work on.

    2sin^{2}(x)+5cos(x)=4where 0 \le x\le360

    AND

    sin(2x)=2cos^{2}(x)where 0\le x\le 2PI
    for the first equation, change \sin^2{x} to 1-\cos^2{x} , move everything to one side, and factor the resulting quadratic.

    for the second, use the double angle identity for sine, move everything to one side, and factor.
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  3. #3
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    I got x = 60 for the first problem, but for the second, when I
    use the double angle identity i get
    2sin(x)cos(x)=2cos^{2}(x)
    i move it
    0=2cos^{2}(x)-2sin(x)cos(x)
    I factor cos(x) out
    0=cos(x)(2cos(x)-2sin(x)
    i factor the 2 out
    2cos(x)(cos(x)-sin(x))
    but then I dont know what to do
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  4. #4
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    Quote Originally Posted by spycrab View Post
    I got x = 60 for the first problem, but for the second, when I
    use the double angle identity i get
    2sin(x)cos(x)=2cos^{2}(x)
    i move it
    0=2cos^{2}(x)-2sin(x)cos(x)
    I factor cos(x) out
    0=cos(x)(2cos(x)-2sin(x)
    i factor the 2 out
    2cos(x)(cos(x)-sin(x))
    but then I dont know what to do
    set each factor equal to 0 ...

    2\cos{x} = 0

    \cos{x} - \sin{x} = 0

    solve each.
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  5. #5
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    im having problems with cosx-sinx = 0. for the 2cosx = 0 part, i get pi/2,
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  6. #6
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    Quote Originally Posted by spycrab View Post
    im having problems with cosx-sinx = 0. for the 2cosx = 0 part, i get pi/2,
    \cos{x} = 0 has two solutions

    \cos{x} = \sin{x} has two solutions

    look at the unit circle ...

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  7. #7
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    Quote Originally Posted by spycrab View Post
    im having problems with cosx-sinx = 0. for the 2cosx = 0 part, i get pi/2,
    Cos(x)=Sin(x)


    For what angles are the horizontal (Cos(x)) and vertical (Sin(x)) co-ordinates of a point on the unit-circle identical ?

    Also, there is a second solution for 2Cosx=0.
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  8. #8
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    Quote Originally Posted by spycrab View Post
    I got x = 60 for the first problem
    you're missing a solution
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  9. #9
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    Thanks!

    I got x=\frac{\pi}{2},x=\frac{\pi}{4},x=\frac{5\pi}{4},x  =\frac{3\pi}{2}
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