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Câu a)$PT \Leftrightarrow 12\left(\sin x-\cos x\right)-12-\sin2x=0$ $\Leftrightarrow 12(\sin x-\cos x) +1-\sin2x -13=0$ $\Leftrightarrow \left(\sin^2x+\cos^2x-sin2x\right)+12\left(\sin x-\cos x\right)-13=0$ $\Leftrightarrow \left(\sin x-\cos x\right)^2+12\left(\sin x-\cos x\right)-13=0$ $\Leftrightarrow \left[\begin{array}{1} \sin x-\cos x=-13\,\,\mbox{(loại)}\\\sin x-\cos x=1 \end{array}\right.$ $\Leftrightarrow \left[\begin{array}{1}x=\dfrac{\pi}{2}+k2\pi \\ x=\pi+k2\pi \end{array}\right.$

Câu a)$PT \Leftrightarrow 12\left(\sin x-\cos x\right)-12-\sin2x=0$ $\Leftrightarrow 12(\sin x-\cos x) +1-\sin2x -13=0$ $\Leftrightarrow \left(\sin^2x+\cos^2x-sin2x\right)+12\left(\sin x-\cos x\right)-13=0$ $\Leftrightarrow \left(\sin x-\cos x\right)^2+12\left(\sin x-\cos x\right)-13=0$ $\Leftrightarrow \left[\begin{array}{1} \sin x-\cos x=-13\,\,\mbox{(loại)}\\\sin x-\cos x=1 \end{array}\right.$ $\Rightarrow \left[\begin{array}{1}x=\dfrac{\pi}{2}+k2\pi \\ x=\pi+k2\pi \end{array}\right.$

Câu a)$PT \Leftrightarrow 12\left(\sin x-\cos x\right)-12-\sin2x=0$ $\Leftrightarrow 12(\sin x-\cos x) +1-\sin2x -13=0$ $\Leftrightarrow \left(\sin^2x+\cos^2x-sin2x\right)+12\left(\sin x-\cos x\right)-13=0$ $\Leftrightarrow \left(\sin x-\cos x\right)^2+12\left(\sin x-\cos x\right)-13=0$ $\Leftrightarrow \left[\begin{array}{1} \sin x-\cos x=-13\,\,\mbox{(loại)}\\\sin x-\cos x=1 \end{array}\right.$ $\Leftrightarrow \left[\begin{array}{1}x=\dfrac{\pi}{2}+k2\pi \\ x=\pi+k2\pi \end{array}\right.$

Câu a)$PT \Leftrightarrow 12\left(\sin x-\cos x\right)-12-\sin2x=0$ $\Leftrightarrow 12(\sin x-\cos x) +1-\sin2x -13=0$ $\Leftrightarrow \left(\sin^2x+\cos^2x-sin2x\right)+12\left(\sin x-\cos x\right)-13=0$ $\Leftrightarrow \left(\sin x-\cos x\right)^2+12\left(\sin x-\cos x\right)-13=0$ $\Leftrightarrow \left[\begin{array}{1} \sin x-\cos x=-13\,\,\mbox{(loại)}\\\sin x-\cos x=1 \end{array}\right.$ $\Leftrightarrow \left[\begin{array}{1}x=\dfrac{\pi}{2}+k2\pi \\ x=\pi+k2\pi \end{array}\right.$

Câu a)$PT \Leftrightarrow 12(sinx-cosx)-12-sin2x=0$$\Leftrightarrow 12(sinx-cosx) +1-sin2x -13=0$$\Leftrightarrow (sin^2x+cos^2x-sin2x)+12(sinx-cosx)-13=0$$\Leftrightarrow (sinx-cosx)^2+12(sinx-cosx)-13=0$$\Leftrightarrow sinx-cosx=-13 (loại)$Hoặc $sinx-cos =1$$\Rightarrow x=\frac{\pi}2+2k\pi$hoặc $x=\pi+2k\pi$

Câu a)$PT \Leftrightarrow 12\left(\sin x-\cos x\right)-12-\sin2x=0$ $\Leftrightarrow 12(\sin x-\cos x) +1-\sin2x -13=0$ $\Leftrightarrow \left(\sin^2x+\cos^2x-sin2x\right)+12\left(\sin x-\cos x\right)-13=0$ $\Leftrightarrow \left(\sin x-\cos x\right)^2+12\left(\sin x-\cos x\right)-13=0$ $\Leftrightarrow \left[\begin{array}{1} \sin x-\cos x=-13\,\,\mbox{(loại)}\\\sin x-\cos x=1 \end{array}\right.$ $\Leftrightarrow \left[\begin{array}{1}x=\dfrac{\pi}{2}+k2\pi \\ x=\pi+k2\pi \end{array}\right.$