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Math Help - Solutions to trig equations

  1. #1
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    Solutions to trig equations

    a)
    b)
    Last edited by mr fantastic; August 10th 2010 at 08:19 PM.
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  2. #2
    Senior Member eumyang's Avatar
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    a) \cos x + 1 = \sin x

    Square both sides:
    (\cos x + 1)^2 = \sin^2 x
    \cos^2 x + 2\cos x + 1 = \sin^2 x

    Use the Pythagorean identity on the RHS:
    \cos^2 x + 2\cos x + 1 = 1 - \cos^2 x

    Move everything to the LHS and you'll have a quadratic on cos x. Solve for cos x, and then find x. Can you take it from here?

    b) 2\cos 3x - 1 = 0

    Solve for cos 3x:
    2\cos 3x = 1
    \cos 3x = \frac{1}{2}

    Now ask yourself, the cosine of what angle(s) is 1/2? (You should know; they are all special angles!) The answers you come up with equal to 3x, which means that you'll have to divide each of the answers by 3 to solve for x.
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  3. #3
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    You would have to prove this mathematically, but clearly from the graph, the first equation has solutions \pi \left( {\cfrac{3}{4}\ \pm\ \cfrac{1}{4} \right)\ +\ 2\pi n\ \ \left( n \in \mathbb Z \right)
    For the second, you need cos(3x)=\cfrac{1}{2}. You should be able to do that.
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  4. #4
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    Quote Originally Posted by eumyang View Post
    a) \cos x + 1 = \sin x

    [...]

    Move everything to the LHS and you'll have a quadratic on cos x. Solve for cos x, and then find x. Can you take it from here?
    Eumyang is right, although this method does lead to an extraneous answer, so you have to be careful: we have:

    2cos^2 x + 2 cos x = 0
    cos x \left( cos x + 1 \right) = 0
    cos x = 0\  or\  -1

    However, this would lead to one set of solutions \cfrac{3\pi}{2} + 2\pi n which actually doesn't fit the original equation. The problem arises from the fact that, given the equations

    (1)\ \  LHS \ = \ RHS
    (2)\ \  (LHS)^2 \ = \ (RHS)^2

    we have (1) implies (2), but not (2) implies (1).
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  5. #5
    Senior Member eumyang's Avatar
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    Quote Originally Posted by Mazerakham View Post
    Eumyang is right, although this method does lead to an extraneous answer, so you have to be careful:
    Uhm, I was kind of hoping that RosieLaird would have realized it on her own. That's why I said, "Can you take it from here?", you know. Thanks for giving it away.
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  6. #6
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    Quote Originally Posted by RosieLaird View Post
    a)
    b)
    (a)

    Another way...

    Let (x,y) be a point on the circumference of the unit circle.

    cos\theta+1=sin\theta

    sin\theta-cos\theta=1

    \displaystyle\huge\frac{y}{r}-\frac{x}{r}=1

    r=1\ \Rightarrow\ y-x=1


    By Pythagoras' theorem

    \sqrt{x^2+y^2}=1

    \sqrt{(y-x)^2}=\sqrt{y^2+x^2}

    \sqrt{y^2-2xy+x^2}=\sqrt{y^2+x^2}

    xy=0\ \Rightarrow\ x=0\ or\ y=0


    Test the 4 possible angles

    cos(0)+1=2\ \ne\ sin(0)

    cos\left(90^o\right)+1=1=sin\left(90^o\right)

    cos\left(180^o\right)+1=0=sin\left(180^o\right)

    cos\left(270^o\right)+1=1\ \ne\ sin\left(270^o\right)
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  7. #7
    Member Mathelogician's Avatar
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    Lightbulb

    Note:
    If squaring both sides as eumyang did, you must al last test the obtained x. Because it may that some soloutions are not real!

    -----------------------------
    Also you can use the method of solving first type of classical trig equations...
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