Chủ đề này có 1 trả lời

[caochinduoi112]

$2sin4x+\sqrt{3}=3sin2x+\sqrt{3}cos2x$

[sieumau88]

$2.sin4x + \sqrt{3} = 3.sin2x + \sqrt{3}.cos2x$

 

$\Leftrightarrow sin2x .\left ( 4.cos2x - 3 \right ) + \sqrt{3}. \left ( 1 - cos2x \right ) = 0$

 

$\Leftrightarrow 2.sinx . cosx . \left ( 4.cos2x - 3 \right ) + \sqrt{3}. \ 2 . \ sin^2x = 0$

 

$\Leftrightarrow \left[ \begin{array}{l} 2 . sinx = 0 \\ 2 \ . 2 \ cos2x \ cosx - 3. \ cosx + \sqrt{3}. sinx = 0 \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} x = k \pi \ , \ \left ( k \in \mathbb{Z} \right ) \\ 2 \ . \left ( cos3x + cosx \right ) - 3. \ cosx + \sqrt{3}. sinx = 0 \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} x = k \pi \ , \ \left ( k \in \mathbb{Z} \right ) \\ cos3x = \dfrac{1}{2} \ cosx \ - \ \dfrac{\sqrt{3}}{2} \ sinx \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} x = k \pi \ , \ \left ( k \in \mathbb{Z} \right ) \\ cos3x = cos\left ( \dfrac{\pi}{3} +x\right ) \end{array} \right.$

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