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

Giải đáp:

\(S = \left\{\dfrac{7\pi}{24} + k\pi;\ \dfrac{13\pi}{24} + k\pi\ \Bigg|\ k\in\Bbb Z\right\}\) 

Lời giải và giải thích chi tiết:

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