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Solvin trig equation

  1. Oct 9, 2005 #1
    I just learned at class how to solve trig equations in this form:

    but i didnt understand ANYTHING!!:eek:
    if someonce can tell me please how to solve equations like that! ( -infinity < x < infinity)

    thank you again,
  2. jcsd
  3. Oct 9, 2005 #2


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    One simple trick is to let a=fcos(y) and b=fsin(y), where f=sqrt(a^2+b^2) and tan(y)=b/a. Then fcos(x-y)+c=0. You can go from there. Notice that if |c|>|f|, there is no real solution.
  4. Oct 9, 2005 #3


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    Hi, the first thing you should focus on understanding, is the trigonometric relation between the cosine (or sine) of a sum (or difference), and the cosines and sines of the terms of the sum, that is, for example:
    [tex]\cos(u-v)=\cos(u)\cos(v)+\sin(u)\sin(v) (1)[/tex]

    Note that the "u"-variable appears in two terms on your right-hand side, in the cos(u) term and the sin(u) term.

    Suppose you had an trigonometric equation of the following form:
    Would you find that difficult to solve?
    Probably not!

    Go now back to your given equation: acos(x)+bsin(x)+c=0
    What if we could utilize the sum formula (1) in a smart manner so that we could transform our equation into something like (2)?
    This is the rationale behind our solution strategy!

    Do you follow thus far?
  5. Oct 9, 2005 #4
    yes, that s interestin'
  6. Oct 9, 2005 #5
    so yeah, let's take this example:
    cos(2x) + 3sin(2x) - 1 = 0
    so we have f=sqrt(10) and then sqrt(10)cos(x-y)-1=0
    where can i go from here ?!
  7. Oct 10, 2005 #6


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    [tex]\cos(2x) + 3 \sin(2x) - 1 = 0[/tex]
    You then divide both sides by : [itex]\sqrt{1 + 3 ^ 2} = \sqrt{10}[/itex]. So, you have:
    [tex]\frac{1}{\sqrt{10}} \cos(2x) + \frac{3}{\sqrt{10}} \sin(2x) = \frac{1}{\sqrt{10}}[/tex]
    Now let:
    [tex]\cos \alpha = \frac{1}{\sqrt{10}}[/tex], and [tex]\sin \alpha = \frac{3}{\sqrt{10}}[/tex]. You can do this because you know that:
    [tex]\left( \frac{1}{\sqrt{10}} \right) ^ 2 + \left( \frac{3}{\sqrt{10}} \right) ^ 2 = 1[/tex], so it's true that [tex]\cos ^ 2 \alpha + \sin ^ 2 \alpha = 1[/tex]
    So you have:
    [tex]\cos \alpha \cos(2x) + \sin \alpha \sin(2x) = \cos \alpha[/tex]
    [tex]\Leftrightarrow \cos(\alpha - 2x) = \cos \alpha[/tex]
    Can you go from here?
    Or you can do it a little bit differently:
    [tex]\cos (2x) + 3 \sin (2x) - 1 = 0[/tex]
    [tex]\Leftrightarrow 1 - 2 \sin ^ 2 x + 6 \sin x \cos x - 1 = 0[/tex]
    [tex]\Leftrightarrow - 2 \sin ^ 2 x + 6 \sin x \cos x = 0[/tex]
    [tex]\Leftrightarrow - \sin ^ 2 x + 3 \sin x \cos x = 0[/tex]
    [tex]\Leftrightarrow \sin x (3 \cos x - \sin x) = 0[/tex]
    Can you go from here?
    Viet Dao,
    Last edited: Oct 10, 2005
  8. Oct 10, 2005 #7
    hmmm yeah!
    thank you!

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