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Trig question

  1. Jun 27, 2005 #1
    solve for

    cos 3x=sin 2x

    using complementary angles and all i got alpha=2x.

    The thing is, if i got alpha 2x from sin inverse, the general formula would be

    [tex] \theta= n \pi + (-1)^n \alpha [/tex]

    i get something like,

    so,....the problem is that i will have x in the general formuala, something i don't want.

    I think my way of doing it is totally wrong, can anyone help?>

    btw, i know that i can use triple and double angle formula, but i've also been taught that inversing sines both sides would also work, but i've forgotten how to do it. Thanks.
    Last edited: Jun 27, 2005
  2. jcsd
  3. Jun 27, 2005 #2


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    I'd write

    [tex]\cos 3x=\sin\left(3x+\frac{\pi}{2}\right) [/tex]

    and then

    [tex] \sin\left(3x+\frac{\pi}{2}\right)-\sin 2x =0 [/tex]

    and then i'd use a trig.identity and end up with 2 simple eqns.

  4. Jun 27, 2005 #3
    Well, as you sould know,


    therefore, your equation would be

    [tex]sin(2\alpha)=sin(\frac{\pi}{2}-3\alpha) [/tex]

    Try it this way...
  5. Jun 27, 2005 #4
    It seems we have posted at the same time :)

    anyway I wouldn't get things into

    [tex] \sin\left(3x+\frac{\pi}{2}\right)-\sin 2x =0 [/tex]

    From what I have written there are only two possibilities. Either




    Maybe you meant just the same but I didn't think they could be the same...
  6. Jun 27, 2005 #5


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    [tex] \sin 2x = \cos 3x [/tex]

    [tex] => \cos (\frac{\pi}{2} - 2x) = \cos 3x [/tex]

    [tex] => \frac{\pi}{2} - 2x = 2n \pi + 3x -I [/tex]


    [tex] \frac{\pi}{2} - 2x = 2n \pi - 3x - II[/tex]

    From this you can solve for the principle value of x by putting n as 0 in I.
  7. Jun 28, 2005 #6
    hey sin[x] equals cos[90-x],so in this case sin[2x] equals cos[90-2x].so equate cos[3x] to cos[90-2x].you should get 18 i guess
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