Another trig equation question

**mp3qz** · July 22nd 2009, 01:02 AM

Let's continue
$sin(2x-\frac{\pi}{6})+4sinx+1=0$

**yeongil** · July 22nd 2009, 04:35 AM

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

01

**mp3qz** · July 22nd 2009, 05:56 AM

If you can, why don't you solve?

**mr fantastic** · July 22nd 2009, 05:59 AM

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?

**mp3qz** · July 23rd 2009, 05:27 AM

sorry, that's my teacher's test.

i can't solve, and this is solution
$sin 2x. \frac{\sqrt[2]{3}}{2} - \frac{1}{2} cos2x+4sinx+1=0$
$2\sqrt[2]{3} sinxcosx-1+ 2sin^2 (x)+2=0$
$2\sqrt[2]{3} sinxcosx+ 2 sin^2 (x)+ sin^2 (x)+ cos^2 (x)=0$

**red_dog** · July 23rd 2009, 08:19 AM

It's something wrong in the second equation. Where is $4\sin x$ ?

**mp3qz** · July 23rd 2009, 07:29 PM

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

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