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Trig question

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:

dextercioby

Science Advisor
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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.

Daniel.
 
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Well, as you sould know,

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

therefore, your equation would be

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

Try it this way...
 
111
0
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

[tex]2\alpha=\frac{\pi}{2}[/tex]

or

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

Maybe you meant just the same but I didn't think they could be the same...
 

siddharth

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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]

or

[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.
 
M

mathelord

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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