a) $4(\sin 3x-\cos 2x)=5(\sin x-1)$
$\Rightarrow 4[3\sin x-4{\sin}^3x-(1-2{\sin}^2x)]=5\sin x-5$
$\Rightarrow 4(3\sin x-4{\sin}^3x-1+2{\sin}^2x)=5\sin x-5$
$\Rightarrow 12\sin x-16{\sin}^3x-4+8{\sin}^2x=5\sin x-5$
$\Rightarrow 16{\sin}^3x-8{\sin}^2x-7\sin x-1=0$
$\Rightarrow \left[\begin{array}{l} \sin x=1 \\ \sin x=\dfrac{-1}{4} \end{array} \right .$
$\Rightarrow \left[ \begin{array}{l} x=\dfrac{\pi}{2}+k2\pi \\ \left[\begin{array}{l} x=\arcsin(\dfrac{-1}{4})+k2\pi \\ x=\pi-\arcsin(\dfrac{-1}{4})+k2\pi\end{array} \right . \end{array} \right .(k\in\mathbb Z)$.
b) $2\sin2x-\cos2x=7\sin x+2\cos x-4$
$\Rightarrow 4\sin x\cos x-1+2{\sin}^2x=7\sin x+2\cos x-4$
$\Rightarrow 2{\sin}^2x-7\sin x+3+4\sin x\cos x-2\cos x=0$
$\Rightarrow (2\sin x-1)(\sin x-3)+2\cos x(2\sin x-1)=0$
$\Rightarrow (2\sin x-1)(\sin x-3+2\cos x)=0$
$\Rightarrow \left[\begin{array}{l} 2\sin x-1=0 \\ \sin x-3+2\cos x=0\end{array} \right .$
$\Rightarrow \left[\begin{array}{l} \sin x=\dfrac{1}{2}(1) \\ \sin x+2\cos x=3(2)\end{array} \right .$
$(1)\Rightarrow \left[\begin{array}{l} x=\dfrac{\pi}{6}+k2\pi \\ x=\dfrac{5\pi}{6}+k2\pi\end{array} \right .(k\in\mathbb Z)$
$(2)\Rightarrow \dfrac{1}{\sqrt5}\sin x+\dfrac{2}{\sqrt5}\cos x=\dfrac{3}{\sqrt5}$0
Đặt $\cos \alpha=\dfrac{1}{\sqrt5}$, $\sin \alpha=\dfrac{2}{\sqrt5}$
$\Rightarrow \cos \alpha\sin x+\sin\alpha\cos x=\dfrac{3}{\sqrt5}$
$\Rightarrow \sin(x+\alpha)=\dfrac{3}{\sqrt5}>1$ (loại).
