Giải phương trình: $2sin(x+\frac{\pi }{3})-sin(2x-\frac{\pi }{6})=\frac{1}{2}$.
GPT: $2sin(x+\frac{\pi }{3})-sin(2x-\frac{\pi }{6})=\frac{1}{2}$.
Started By axe900, 20-06-2013 - 00:39
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Posted 21-06-2013 - 21:38
sieumau88
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pt $\Leftrightarrow 2. \ sin\left ( x+\dfrac{\pi }{3} \right ) + sin\left ( -2x+\dfrac{\pi }{6} \right ) - \ \dfrac{1}{2} = 0$
$\Leftrightarrow 2. \ sin\left ( x+\dfrac{\pi }{3} \right ) + cos\left ( 2x+\dfrac{\pi }{3} \right ) - \ cos\dfrac{\pi}{3} = 0$
$\Leftrightarrow 2. \ sin\left ( x+\dfrac{\pi }{3} \right ) - 2 . \ sin\left ( x+\dfrac{\pi }{3} \right ) \ . sinx = 0$
$\Leftrightarrow 2. \ sin\left ( x+\dfrac{\pi }{3} \right ) \ . \left (1 - sinx \right ) = 0$
$\Leftrightarrow \left[ \begin{array}{l} sin\left ( x+\dfrac{\pi }{3} \right ) = 0 \\ sinx = 1 \end{array} \right.$
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$\Leftrightarrow 2. \ sin\left ( x+\dfrac{\pi }{3} \right ) + cos\left ( 2x+\dfrac{\pi }{3} \right ) - \ cos\dfrac{\pi}{3} = 0$
$\Leftrightarrow 2. \ sin\left ( x+\dfrac{\pi }{3} \right ) - 2 . \ sin\left ( x+\dfrac{\pi }{3} \right ) \ . sinx = 0$
$\Leftrightarrow 2. \ sin\left ( x+\dfrac{\pi }{3} \right ) \ . \left (1 - sinx \right ) = 0$
$\Leftrightarrow \left[ \begin{array}{l} sin\left ( x+\dfrac{\pi }{3} \right ) = 0 \\ sinx = 1 \end{array} \right.$
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