Đáp án:

\(\begin{array}{l}
a)\quad S =\left\{-\dfrac{\pi}{6} + k\pi;\ \dfrac{\pi}{3} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}\\
b)\quad S =\left\{- \dfrac{\pi}{3}  +k2\pi;\ - \dfrac{\pi}{9} + k\dfrac{2\pi}{3}\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
a)\quad \sin x - \sqrt3\cos x =2\sin3x\\
\Leftrightarrow \dfrac12\sin x - \dfrac{\sqrt3}{2}\cos x = \sin3x\\
\Leftrightarrow \sin\left(x - \dfrac{\pi}{3}\right) = \sin3x\\
\Leftrightarrow \left[\begin{array}{l}x - \dfrac{\pi}{3} = 3x + k2\pi\\x - \dfrac{\pi}{3} = \pi - 3x + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x =-\dfrac{\pi}{6} + k\pi\\x = \dfrac{\pi}{3} + k\dfrac{\pi}{2}\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S =\left\{-\dfrac{\pi}{6} + k\pi;\ \dfrac{\pi}{3} + k\dfrac{\pi}{2}\ \Bigg|\ k\in\Bbb Z\right\}\\
b)\quad \cos2x - \sqrt3\sin2x - 2\cos x = 0\\
\Leftrightarrow \dfrac12\cos2x- \dfrac{\sqrt3}{2}\sin2x = \cos x\\
\Leftrightarrow \cos\left(2x + \dfrac{\pi}{3}\right) =\cos x \\
\Leftrightarrow \left[\begin{array}{l}2x + \dfrac{\pi}{3} = x + k2\pi\\2x + \dfrac{\pi}{3} = -x + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = - \dfrac{\pi}{3}  +k2\pi\\x = - \dfrac{\pi}{9} + k\dfrac{2\pi}{3}\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy}\ S =\left\{- \dfrac{\pi}{3}  +k2\pi;\ - \dfrac{\pi}{9} + k\dfrac{2\pi}{3}\ \Bigg|\ k\in\Bbb Z\right\}
\end{array}\) 

`a)``sinx-\sqrt{3}cosx=2sin3x`

`->1/2 sinx -\sqrt{3}/2 cosx=sin2x`

`->sin(x-\pi/3)=sin3x`

`->`\(\left[ \begin{array}{l}x-\frac{\pi}{3}=3x+k2\pi\\x-\frac{\pi}{3}=\pi-3x+k2\pi\end{array} \right.\) 

`->`\(\left[ \begin{array}{l}x=-\frac{\pi}{6}+k\pi\\x=\frac{\pi}{3}+k\frac{\pi}{2}\end{array} \right.\) 

`->x=\pi/3+k \pi/2(k∈Z)`

`b)``cos2x+\sqrt{3}sin2x-2cosx=0`

`->1/2 cos2x-\sqrt{3}/2 sin2x-cosx=0`

`->cos \pi/3 cos2x-sin \pi/3 sin 2x = cosx`

`->cos(2x+\pi/3)=cosx`

`->`\(\left[ \begin{array}{l}2x+\frac{\pi}{3}=x+k2\pi\\2x+\frac{\pi}{3}=-x+k2\pi\end{array} \right.\) 

`->`\(\left[ \begin{array}{l}x=-\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{9}+k\frac{2\pi}{3}\end{array} \right.\) `(k∈Z)`