Bài tập nâng cao Phương trình lượng giác cơ bản · Bài 80

Điều kiện: $\cos x.\sin y\neq 0$
$\left\{ \begin{array}{l} \sin x.\cos y=\frac{-1}{2} \\ \tan x.\cot y=1 \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} \frac{1}{2}\left[ \sin\left( x+y \right)+\sin\left( x-y \right) \right]=\frac{-1}{2} \\ \frac{\sin x.\cos y}{\cos x.\sin y}-1=0 \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} \sin\left( x+y \right)+\sin\left( x-y \right)=-1 \\ \sin x\cos y-\sin y.\cos x=0 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} \sin\left( x+y \right)+\sin\left( x-y \right)=-1 \\ \sin\left( x-y \right)=0 \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} \sin\left( x+y \right)=-1 \\ \sin\left( x-y \right)=0 \end{array} \right.$ $\Leftrightarrow \left\{ \begin{array}{l} x+y=-\frac{\pi}{2}+k2\pi \\ x-y=h\pi \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} x=-\frac{\pi}{4}+\left( 2k+h \right)\frac{\pi}{2} \\ y=\frac{-\pi}{4}+\left( 2k-h \right)\frac{\pi}{2} \end{array} \right.$