# Math Help - Another trig equation question

1. ## Another trig equation question

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

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

01

3. If you can, why don't you solve?

4. 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?

5. sorry, that's my teacher's test.

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

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

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