Let's continue $\displaystyle sin(2x-\frac{\pi}{6})+4sinx+1=0$
Last edited by mp3qz; Jul 22nd 2009 at 05:53 AM.
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Start by using the sine of a difference formula: $\displaystyle \sin (u - v) = \sin u \cos v - \cos u \sin v$. 01
Last edited by mr fantastic; Jul 22nd 2009 at 05:00 AM.
If you can, why don't you solve?
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?
sorry, that's my teacher's test. i can't solve, and this is solution
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
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