Let's continue
$\displaystyle sin(2x-\frac{\pi}{6})+4sinx+1=0$
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Let's continue
$\displaystyle sin(2x-\frac{\pi}{6})+4sinx+1=0$
Start by using the sine of a difference formula:
$\displaystyle \sin (u - v) = \sin u \cos v - \cos u \sin v$.
01
If you can, why don't you solve?
THis is MathHelpForum, not MathSolvetheproblemcompletelyForum.Quote:
Originally Posted by mp3qz
You've been given a suggestion. Have you tried it? How far did you get? Where are you stuck?
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
It's something wrong in the second equation. Where is $\displaystyle 4\sin x$?
sorry, it's wrong.
sorry every body. But who can help me this equation? PLEASE(Worried)
