Lời giải:
$4(\sin 3x-\cos 2x)=5(\sin x+1)$
$\Leftrightarrow 4[\sin 2x\cos x-\cos 2x\sin x-\cos 2x)=5(\sin x+1)$
$\Leftrightarrow 4\sin 2x\cos x-4\cos 2x(\sin x+1)=5(\sin x+1)$
$\Leftrightarrow 4\sin 2x\cos x-(4\cos 2x+5)(\sin x+1)=0$
$\Leftrightarrow 8\sin x\cos ^2x-(4\cos 2x+5)(\sin x+1)=0$
$\Leftrightarrow 8\sin x(1-\sin ^2x)-(4\cos 2x+5)(\sin x+1)=0$
$\Leftrightarrow (\sin x+1)[8\sin x(1-\sin x)-(4\cos 2x+5)]=0$
Nếu $\sin x+1=0$
$\Leftrightarrow \sin x=-1\Rightarrow x=\frac{-\pi}{2}+2k\pi$
Nếu $8\sin x(1-\sin x)-(4\cos 2x+5)=0$
$\Leftrightarrow 8\sin x-8\sin ^2x=4(1-2\sin ^2x)+5$
$\Leftrightarrow 8\sin x=9$
$\Leftrightarrow \sin x=\frac{9}{8}>1$ (loại)
Vậy........
