Let's continue

$\displaystyle sin(2x-\frac{\pi}{6})+4sinx+1=0$

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- Jul 22nd 2009, 01:02 AMmp3qzAnother trig equation question
Let's continue

$\displaystyle sin(2x-\frac{\pi}{6})+4sinx+1=0$ - Jul 22nd 2009, 04:35 AMyeongil
Start by using the sine of a difference formula:

$\displaystyle \sin (u - v) = \sin u \cos v - \cos u \sin v$.

01 - Jul 22nd 2009, 05:56 AMmp3qz
If you can, why don't you solve?

- Jul 22nd 2009, 05:59 AMmr fantastic
- Jul 23rd 2009, 05:27 AMmp3qz
sorry, that's my teacher's test.

i can't solve, and this is solution

http://dientuvietnam.net/cgi-bin/mim...os2x+4sinx+1=0

http://dientuvietnam.net/cgi-bin/mim...2sin^2 (x)+2=0

http://dientuvietnam.net/cgi-bin/mim...)+ cos^2 (x)=0 - Jul 23rd 2009, 08:19 AMred_dog
It's something wrong in the second equation. Where is $\displaystyle 4\sin x$?

- Jul 23rd 2009, 07:29 PMmp3qz
sorry, it's wrong.

sorry every body. But who can help me this equation? PLEASE(Worried)