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Math Help - Solving Trig Equations

  1. #1
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    Solving Trig Equations

    Solve each equation for  0 \leq \theta \leq 360


    --
    sin^2x - 1 = cos^2x
    --

    Work:
    cos^2x - cos^2x = 0
    0 = 0


    --
    Textbook Answers:
    90 and 270
    --




    Incorporate the use of the double angle formulas identities and other identities to solve the following on  0 \leq \theta \leq 360

    --
    sin2x = 2cos2x
    --

    Work:
    2sinxcosx - 2cosx = 0
    2cosx (sinx - 1) = 0

    What do I do now?

    --
    Textbook Answers:
    90 and 270
    --



    Btw, how would you simply  cos^2x to  cosx if you had something like cos^2x = 1? Would it be...
    cosx = \sqrt{1}
     x = cos^-1 \sqrt{1}?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    --
    sin^2x - 1 = cos^2x
    --

    Work:
    cos^2x - cos^2x = 0
    0 = 0
    Not quite. 1 - sin^2(x) = cos^2(x), so
    sin^2x - 1 = cos^2x

    -cos^2(x) = cos^2(x)

    2cos^2(x) = 0

    cos(x) = 0

    Thus x = 90^o , 270^o

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    Incorporate the use of the double angle formulas identities and other identities to solve the following on  0 \leq \theta \leq 360

    --
    sin2x = 2cos2x
    --

    Work:
    2sinxcosx - 2cosx = 0
    2cosx (sinx - 1) = 0

    What do I do now?
    You forgot about the cos(2x).

    And check your equation. Your given solutions don't work.

    This is one of the nastier ones:
    sin(2x) = 2cos(2x)

    2sin(x)cos(x) = 2cos(2x)

    sin(x) cos(x) = cos(2x)

    sin(x) cos(x) = 2cos^2(x) - 1

    cos(x) (\pm)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1

    cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1

    5cos^4(x) - 5cos^2(x) + 1 = 0

    This has solutions
    cos(x) = \pm 0.85065080835215 \implies x = 31.7175^o, 148.283^0
    and
    cos(x) = \pm 0.52573111211905 \implies x = 58.2825^o, 121.717^o

    Of these, only x = 31.7175^o, 121.717^o solve the original equation.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    Btw, how would you simply  cos^2x to  cosx if you had something like cos^2x = 1? Would it be...
    cosx = \sqrt{1}
     x = cos^-1 \sqrt{1}?
    How do you solve any equation where you take the square root of both sides?

    cos^2(x) = 1

    cos(x) = \pm 1

    Thus
    x = cos^{-1}(1) = 0^o
    and
    x = cos^{-1}(-1) = 180^o

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    You forgot about the cos(2x).

    And check your equation. Your given solutions don't work.

    This is one of the nastier ones:
    sin(2x) = 2cos(2x)

    2sin(x)cos(x) = 2cos(2x)

    sin(x) cos(x) = cos(2x)

    sin(x) cos(x) = 2cos^2(x) - 1

    cos(x) (\pm)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1

    cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1

    5cos^4(x) - 5cos^2(x) + 1 = 0

    This has solutions
    cos(x) = \pm 0.85065080835215 \implies x = 31.7175^o, 148.283^0
    and
    cos(x) = \pm 0.52573111211905 \implies x = 58.2825^o, 121.717^o

    Of these, only x = 31.7175^o, 121.717^o solve the original equation.

    -Dan
    What the heck...? I'm so confused
    I'm pretty sure this equation isn't that complicated...
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  6. #6
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    Quote Originally Posted by topsquark View Post
    Not quite. 1 - sin^2(x) = cos^2(x), so
    sin^2x - 1 = cos^2x

    -cos^2(x) = cos^2(x)

    2cos^2(x) = 0

    cos(x) = 0

    Thus x = 90^o , 270^o

    -Dan
    How did you get -cos^2x? Isn't it cos^2x for sin^2x - 1?
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    What the heck...? I'm so confused
    I'm pretty sure this equation isn't that complicated...
    This equation is. Whatever your book (teacher, professor, whatever) intended for you to solve certainly wasn't given the listed solution set. There's a typo in your given equation, but I can't figure out what it was meant to be.

    -Dan
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Macleef View Post
    How did you get -cos^2x? Isn't it cos^2x for sin^2x - 1?
    sin^2(x) + cos^2(x) = 1

    sin^2(x) + cos^2(x) - 1 = 0

    sin^2(x) - 1 = -cos^2(x)

    -Dan
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