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a) $(2\cos x-1)(2\sin2x+1)+4\sin\frac{3x}{2}\sin\dfrac{x}{2}=1$
$\Leftrightarrow 4\cos x \sin 2x+2\cos x-2\sin 2x-1-2(\cos 2x -\cos x)=1$
$\Leftrightarrow 4\cos x \sin 2x+4\cos x-2\sin 2x-2\cos 2x -2=0$
$\Leftrightarrow 2\cos x (\sin 2x+1)-(\sin 2x+1)-\cos 2x =0$
$\Leftrightarrow 2\cos x (\sin x+\cos x)^2-(\sin x+\cos x)^2-(\sin x+\cos x)(\cos x- \sin x)=0$
$\Leftrightarrow (\sin x+\cos x)(2\cos x\sin x+2\cos^2x-\sin x-\cos x-\cos x+ \sin x)=0$
$\Leftrightarrow (\sin x+\cos x)(\cos x\sin x+\cos^2x-2\cos x)=0$
$\Leftrightarrow (\sin x+\cos x)\cos x(\sin x+\cos x-2)=0$
Đến đây đơn giản em tự giải nốt nhé.
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