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Có $cos(\frac{5\pi}{12}-x)=cos(\frac{6\pi}{12}-\frac{\pi}{12}-x)=sin(x+\frac{\pi}{12})$$PT \Leftrightarrow \sqrt2 . 2sinx.sin(x+\frac{\pi}{12})=1$$\Leftrightarrow \sqrt2 .[cos (2x+\frac{\pi}{12})-cos\frac{\pi}{12}]=1$$\Leftrightarrow \sqrt2.cos(2x+\pi/12)=1+\sqrt2.cos\frac{\pi}{12}=\frac{3+\sqrt3}2$$\Leftrightarrow cos(2x+\pi/12)=\frac{\sqrt6+3\sqrt2}4 >1$Vậy PT vô nghiệm

Có $cos(\frac{5\pi}{12}-x)=cos(\frac{6\pi}{12}-\frac{\pi}{12}-x)=sin(x+\frac{\pi}{12})$$PT \Leftrightarrow \sqrt2 . 2sinx.sin(x+\frac{\pi}{12})=1$$\Leftrightarrow -\sqrt2 .[cos (2x+\frac{\pi}{12})-cos\frac{\pi}{12}]=1$$\Leftrightarrow -\sqrt2.cos(2x+\pi/12)=1-\sqrt2.cos\frac{\pi}{12}=\frac{1-\sqrt3}2$$\Leftrightarrow cos(2x+\pi/12)=\frac{\sqrt6-\sqrt2}4=cos(\pm \frac{5\pi}{12}) $$x=\frac{\pi}6$ hoặc $x=-\frac{\pi}4$