Another trig equation question

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Jul 22nd 2009, 01:02 AM
mp3qz

Another trig equation question

Let's continue
$sin(2x-\frac{\pi}{6})+4sinx+1=0$
Jul 22nd 2009, 04:35 AM
yeongil

Start by using the sine of a difference formula:
$\sin (u - v) = \sin u \cos v - \cos u \sin v$ .

01
Jul 22nd 2009, 05:56 AM
mp3qz

If you can, why don't you solve?
Jul 22nd 2009, 05:59 AM
mr fantastic

Quote:

Originally Posted by mp3qz

If you can, why don't you solve?

THis is MathHelpForum, not MathSolvetheproblemcompletelyForum.

You've been given a suggestion. Have you tried it? How far did you get? Where are you stuck?
Jul 23rd 2009, 05:27 AM
mp3qz

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 AM
red_dog

It's something wrong in the second equation. Where is $4\sin x$ ?
Jul 23rd 2009, 07:29 PM
mp3qz

sorry, it's wrong.
sorry every body. But who can help me this equation? PLEASE(Worried)

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