1.$sin8x - 4cos4x=0$
   2.$\frac{sin2x}{cosx}+tanxcosx-2sinx+1=0$
   3.$1+ sinx = cos\frac{x}{2}$
     
      Mọi người giúp đỡ nhé ^^ 
 
3. 1+sinx = cosx/2 
\Leftrightarrow  1+ sinx = sin(x/2 + \Pi/2)
\Leftrightarrow 1+ sinx = sin x/2 + 1
\Leftrightarrow sinx      = sin\Pi/2
\Leftrightarrow x = x/2 + k2\Pi    hoặc x= \Pi - x/2 + k2\Pi
\Leftrightarrow x = x/2 + k2\Pi    hoặc x= 2\Pi /3 + k4\Pi /3

 1 , $sin8x-4cos4x=0$
    $\Leftrightarrow 2sin4x.cos4x-4cos4x=0$
    $\Leftrightarrow 2cos4x(sin4x-2)=0$
    $\Leftrightarrow cos4x=0$ hoặc $sin4x=2$  ( loại sin4x=2 vì 2>1)
 2, $\frac{sin2x}{cosx}+tanxcosx-2sinx+1=0$   ( ĐK : $cosx\neq 0)$
    $\Leftrightarrow 2sinxcosx+tanx.cos^{2}x-2sinxcosx+cosx=0$
    $\Leftrightarrow cosx(sinx+1)=0$
    $\Leftrightarrow sinx=-1$  hoặc $cosx=0$ (loại $cosx=0)$
 
uk nhầm . –  Dân Nguyễn 07-06-14 07:42 AM
quy đồng còn thiếu kìa Death êi –  Minn 06-06-14 11:33 PM
Làm sai bài cmn 2 =)) –  Nero 06-06-14 11:17 PM
ghét nhất nghĩ ra đc mà lười gõ CT :)) –  Minn 06-06-14 11:14 PM
chú này nhah ghê –  Minn 06-06-14 11:14 PM
bạn làm bài 2 thông minh hơn mình tưởng :D –  Nero 06-06-14 11:02 PM

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