1. sin x - cos x $\frac{1}{cos x}$= $\frac{1}{sin x}$ (1)
Đk: sin x $\neq$ 0, cos x $\neq$ 0
pt(1) ⇔ sin²x cosx -cos²x sinx +sin x -cos x =0 ⇔sinx cosx ( sin x - cos x) +(sin x -cos x) = 0 ⇔ (sin x -cos x)*(sin x cos x +1)=0 ⇔ \(\left[ \begin{array}{l}sin x=cos x\\\frac{1}{2}sin2x=-1\end{array} \right.\) ⇔\(\left[ \begin{array}{l}cos x =cos(\frac{\pi}{2}-x)\\sin2x=-2(ktm)\end{array} \right.\) ⇔ \(\left[ \begin{array}{l} x=\frac{\pi}{2}-x+k2\pi\\x=-\frac{\pi}{2}+x+k2\pi(L)\end{array}\right.\) (k∈Z) ⇔ x=$\frac{\pi}{4}$ +k$\pi$ (k∈Z) Vậy ...
