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PT
$\Leftrightarrow \left | \cos x \right |-\left | \sin x \right |=\cos2x.\sqrt{1+\sin2x}$
$\Leftrightarrow \left | \cos x \right |-\left | \sin x \right |=\left ( \left | \cos x \right |-\left | \sin x \right | \right )\left ( \left | \cos x \right |+\left | \sin x \right | \right )\sqrt{1+\sin2x}$
$\Leftrightarrow \left[ {\begin{matrix} \left | \cos x \right |=\left | \sin x \right |\qquad (1)\\ \left ( \left | \cos x \right |+\left | \sin x \right | \right ) \sqrt{1+\sin2x}=1\qquad (2)\end{matrix}} \right.$
+ Giải $(1)$
$(1)\Leftrightarrow \cos^2 x =\sin^2 x \Leftrightarrow \cos 2x =0\Leftrightarrow x =\dfrac{\pi}{4}+k\dfrac{\pi}{2},\quad k \in \mathbb Z.$
Từ $\begin{cases} x =\dfrac{\pi}{4}+k\dfrac{\pi}{2} \\ 2007 <x <2008 \\k \in \mathbb Z\end{cases}\Rightarrow $ không tồn tại $k,x$
+ Giải $(2)$
$(2)\Leftrightarrow (1+|\sin2x|)(1+\sin2x)=1 \Leftrightarrow \sin 2x =0\Leftrightarrow x =k\dfrac{\pi}{2},\quad k \in
\mathbb Z.$
Từ $\begin{cases} x =k\dfrac{\pi}{2} \\
2007 <x <2008 \\k \in \mathbb Z\end{cases}\Rightarrow \boxed{x=639 \pi}.$
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