Phương pháp đặt:

$\begin{array}{l} \sin x - \cos x = t\\  \Leftrightarrow \sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) = t\\  \Rightarrow t \in \left[ { - \sqrt 2 ;\sqrt 2 } \right]\\  \Rightarrow {t^2} = 1 - 2\sin x\cos x = 1 - \sin 2x\\  \Leftrightarrow \sin 2x = 1 - {t^2} \end{array}$

2)

$\begin{array}{l} 5\sin 2x - 12\left( {\sin x - \cos x} \right) + 12 = 0\\  \Leftrightarrow 5\left( {1 - {t^2}} \right) - 12t + 12 = 0\\  \Leftrightarrow 5{t^2} + 12t - 17 = 0\\  \Leftrightarrow \left[ \begin{array}{l} t = 1\\ t =  - \dfrac{{17}}{5}(L) \end{array} \right.\\  \Rightarrow \sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) = 1\\  \Leftrightarrow \sin \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\\  \Leftrightarrow \left[ \begin{array}{l} x - \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\ x - \dfrac{\pi }{4} = \dfrac{{3\pi }}{4} + k2\pi  \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = k2\pi \\ x = \pi  + k2\pi  \end{array} \right. \Rightarrow x = k\pi \left( {k \in \mathbb{Z}} \right) \end{array}$

4) Cách làm tương tự

$\begin{array}{l} \cos x - \sin x + 3\sin 2x - 1 = 0\\  \Leftrightarrow  - \left( {\sin x - \cos x} \right) + 3\sin 2x - 1 = 0\\  \Leftrightarrow  - t + 3\left( {1 - {t^2}} \right) - 1 = 0\\  \Leftrightarrow 3{t^2} + t - 2 = 0\\  \Leftrightarrow \left[ \begin{array}{l} t = 1\\ t =  - \dfrac{2}{3} \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} \sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) = 1\\ \sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) =  - \dfrac{2}{3} \end{array} \right.\\  \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ x = \arcsin \left( { - \dfrac{{\sqrt 2 }}{3}} \right) + k2\pi \\ x = \pi  - \arcsin \left( {\dfrac{{ - \sqrt 2 }}{3}} \right) + k2\pi  \end{array} \right.\left( {k \in \mathbb{Z}} \right) \end{array}$