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Thread: Solving Trig Equations

  1. #1
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    Solving Trig Equations

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


    --
    $\displaystyle sin^2x - 1 = cos^2x $
    --

    Work:
    $\displaystyle cos^2x - cos^2x = 0$
    $\displaystyle 0 = 0 $


    --
    Textbook Answers:
    90 and 270
    --




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

    --
    $\displaystyle sin2x = 2cos2x$
    --

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

    What do I do now?

    --
    Textbook Answers:
    90 and 270
    --



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

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

    $\displaystyle -cos^2(x) = cos^2(x)$

    $\displaystyle 2cos^2(x) = 0$

    $\displaystyle cos(x) = 0$

    Thus $\displaystyle x = 90^o , 270^o$

    -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 $\displaystyle 0 \leq \theta \leq 360$

    --
    $\displaystyle sin2x = 2cos2x$
    --

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

    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:
    $\displaystyle sin(2x) = 2cos(2x)$

    $\displaystyle 2sin(x)cos(x) = 2cos(2x)$

    $\displaystyle sin(x) cos(x) = cos(2x)$

    $\displaystyle sin(x) cos(x) = 2cos^2(x) - 1$

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

    $\displaystyle cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1$

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

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

    Of these, only $\displaystyle x = 31.7175^o, 121.717^o$ 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 $\displaystyle cos^2x $ to $\displaystyle cosx$ if you had something like $\displaystyle cos^2x = 1$? Would it be...
    $\displaystyle cosx = \sqrt{1}$
    $\displaystyle x = cos^-1 \sqrt{1}$?
    How do you solve any equation where you take the square root of both sides?

    $\displaystyle cos^2(x) = 1$

    $\displaystyle cos(x) = \pm 1$

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

    -Dan
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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:
    $\displaystyle sin(2x) = 2cos(2x)$

    $\displaystyle 2sin(x)cos(x) = 2cos(2x)$

    $\displaystyle sin(x) cos(x) = cos(2x)$

    $\displaystyle sin(x) cos(x) = 2cos^2(x) - 1$

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

    $\displaystyle cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1$

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

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

    Of these, only $\displaystyle x = 31.7175^o, 121.717^o$ 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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    Quote Originally Posted by topsquark View Post
    Not quite. $\displaystyle 1 - sin^2(x) = cos^2(x)$, so
    $\displaystyle sin^2x - 1 = cos^2x $

    $\displaystyle -cos^2(x) = cos^2(x)$

    $\displaystyle 2cos^2(x) = 0$

    $\displaystyle cos(x) = 0$

    Thus $\displaystyle x = 90^o , 270^o$

    -Dan
    How did you get $\displaystyle -cos^2x$? Isn't it $\displaystyle cos^2x $ for $\displaystyle sin^2x - 1$?
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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 $\displaystyle -cos^2x$? Isn't it $\displaystyle cos^2x $ for $\displaystyle sin^2x - 1$?
    $\displaystyle sin^2(x) + cos^2(x) = 1$

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

    $\displaystyle sin^2(x) - 1 = -cos^2(x)$

    -Dan
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