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PT
$\Leftrightarrow (9\sin x-9) + (6\cos x - 3\sin 2x) + \cos 2x +1= 0$
$\Leftrightarrow 9(\sin x-1) + 6\cos x(1-\sin x) +2(1-\sin^2 x)= 0$
$\Leftrightarrow (\sin x-1)(9 - 6\cos x -2-2\sin x)= 0$
$\Leftrightarrow (\sin x-1)(7 - 6\cos x -2\sin x)= 0$
Mặt khác ta thấy $6^2 +2^2 <7^2$ nên PT $7 = 6\cos x +2\sin x$ vô nghiệm.
Do vậy $\sin x= 1 \Leftrightarrow x = \dfrac{\pi}{2}+k2\pi, k \in \mathbb Z.$
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