# Trigo Eqn

• December 1st 2009, 09:47 PM
gp3
Trigo Eqn
How to solve this to get the angle?

sin6x=3sin3x for 0 < x < 2pi

this is how far i got:

2sin3xcos3x = 3sin3x
2sin3xcos3x/sin3x = 3
2cos3x = 3
cos3x = 3/2
3x = cos-1 (3/2)

But the calculator cant compute, gives me an error. Can someone see where i went wrong?

Thanks!!!!
• December 1st 2009, 10:08 PM
Quote:

Originally Posted by gp3
How to solve this to get the angle?

sin6x=3sin3x for 0 < x < 2pi

this is how far i got:

2sin3xcos3x = 3sin3x
2sin3xcos3x/sin3x = 3
2cos3x = 3
cos3x = 3/2
3x = cos-1 (3/2)

But the calculator cant compute, gives me an error. Can someone see where i went wrong?

Thanks!!!!

HI

you started off well .

$\sin 6x=3\sin 3x$

$2\sin 3x\cos 3x=3\sin 3x$ **

$2\sin 3x\cos 3x-3\sin 3x=0$

$\sin 3x(2\cos 3x-3)=0$

$\sin 3x=0 OR \cos 3x=\frac{2}{3}$

Can you pick it up from here ? Just remember that the range is 0<x<2pi which implies 0<3x<6pi

Note that at ** , you cant cancel off the sin 3x at both sides or else you would be missing solutions because you are ignoring the possibility of
sin 3x=0

Its like $2x^2=x$ , you cannot cancel the x so that x=1/2 , this is wrong .
• December 1st 2009, 10:18 PM
gp3
yea, i know how to carry on. But if I have any more doubts, I will post them here.(Wink) Thanks for your fast reply! (Rofl)
• December 1st 2009, 10:26 PM
gp3
oh yea, i just spotted a mistake, i believe it should be cos3x = 3/2 and not cos3x = 2/3. And thus i cant solve cos3x = 3/2, it gives me an error.. help pls!
• December 2nd 2009, 01:00 AM
Quote:

Originally Posted by gp3
oh yea, i just spotted a mistake, i believe it should be cos3x = 3/2 and not cos3x = 2/3. And thus i cant solve cos3x = 3/2, it gives me an error.. help pls!

oh yeah , my mistake , sorry .

cos 3x=3/2 , it is undefined , because maximum for cos is 1 , anything over it would be undefined so you can just ignore this since it will not yield any solutions .
• December 2nd 2009, 02:42 AM
gp3
hmm...so what i got for my answer is 0, pi/3, 2pi/3.

But the solution they gave me was 0, pi/3, 2pi/3, pi, 4pi/3 and 5pi/3.

tbh, i know the range is something like we have to go 2 rounds? or 3 rounds? for this question but how do you go about doing it? any idea on how to get the final 3 ?
• December 2nd 2009, 02:47 AM
mr fantastic
Quote:

Originally Posted by gp3
hmm...so what i got for my answer is 0, pi/3, 2pi/3.

But the solution they gave me was 0, pi/3, 2pi/3, pi, 4pi/3 and 5pi/3.

tbh, i know the range is something like we have to go 2 rounds? or 3 rounds? for this question but how do you go about doing it? any idea on how to get the final 3 ?

Well, since they say find all solutions for 0 < x < 2pi, that's what you're expected to do. Why did you stop at 2pi/3?

The general solution to $\sin (3x) = 0$ is $3x = n \pi \Rightarrow x = \frac{n \pi}{3}$ where n is an integer. Use this to get all the solutions that lie in 0 < x < 2pi.
• December 2nd 2009, 02:53 AM
gp3
lol....ok i got it now.(Giggle)