Chủ đề này có 1 trả lời

[axe900]

Giải phương trình: $2sin(x+\frac{\pi }{3})-sin(2x-\frac{\pi }{6})=\frac{1}{2}$.

[sieumau88]

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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