2sinx(1+tanx.tan$\frac{x}{2}$)=sin2x+1
Edited by poyvinhhung, 27-06-2013 - 21:46.
2sinx(1+tanx.tan$\frac{x}{2}$)=sin2x+1
Edited by poyvinhhung, 27-06-2013 - 21:46.
$2.sinx.\left ( 1+tanx.tan\dfrac{x}{2} \right )=sin2x+1$
ĐK: $\left\{\begin{matrix}
cos \ x \neq 0 \\
cos \ \dfrac{x}{2} \neq 0
\end{matrix}\right.$
pt $\Leftrightarrow 2.sinx.\left ( 1+\dfrac{sinx.sin\frac{x}{2}}{cosx.cos\frac{x}{2}} \right )=sin2x+1$
$\Leftrightarrow 2.sinx.\left ( \dfrac{cosx.cos\frac{x}{2}+sinx.sin\frac{x}{2}}{cosx.cos\frac{x}{2}} \right )=sin2x+1$
$\Leftrightarrow 2.sinx.\left [ \dfrac{cos\left (x-\frac{x}{2} \right )}{cosx.cos\frac{x}{2}} \right ]=sin2x+1$
$\Leftrightarrow 2.tanx=sin2x+1$
$\Leftrightarrow 2.t = \dfrac{2.t}{t^2+1}+1$ ( Đặt $t=tanx$)
$\Leftrightarrow 2.t+2.t^3 =2.t+t^2+1$
$\Leftrightarrow 2.t^3 -t^2-1=0$ (pt này có 1 nghiệm là t=1)
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