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222-9+11+12:2*14+14 = ? ( )

## Toán Lớp 11: sin3xcosx-$\sqrt{3}$ cos2x=$\sqrt{2}$ +sinxcos3x

Toán Lớp 11: sin3xcosx-$\sqrt{3}$ cos2x=$\sqrt{2}$ +sinxcos3x

$$S = \left\{\dfrac{7\pi}{24} + k\pi;\ \dfrac{13\pi}{24} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}$$
$$\begin{array}{l} \quad \sin3x.\cos x -\sqrt3\cos2x = \sqrt2+\sin x\cos3x\\ \Leftrightarrow (\sin3x.\cos x-\cos3x.\sin x) – \sqrt3\cos2x = \sqrt2\\ \Leftrightarrow \sin2x – \sqrt3\cos2x = \sqrt2\\ \Leftrightarrow \dfrac12\sin2x – \dfrac{\sqrt3}{2}\cos2x= \dfrac{\sqrt2}{2}\\ \Leftrightarrow \sin\left(2x – \dfrac{\pi}{3}\right) = \sin\dfrac{\pi}{4}\\ \Leftrightarrow \left[\begin{array}{l}2x – \dfrac{\pi}{3}=\dfrac{\pi}{4} + k2\pi\\2x – \dfrac{\pi}{3} = \dfrac{3\pi}{4} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{7\pi}{24} + k\pi\\x = \dfrac{13\pi}{24} + k\pi\end{array}\right.\quad (k\in\Bbb Z)\\ \text{Vậy}\ S = \left\{\dfrac{7\pi}{24} + k\pi;\ \dfrac{13\pi}{24} + k\pi\ \Bigg|\ k\in\Bbb Z\right\} \end{array}$$