Giải phương trình sau: $\tan \left ( \frac{\pi }{4}-x \right )$ = $\tan 2x$
Cách 1:
$\tan \left ( \frac{\pi }{4}-x \right )$ = $\tan 2x$
$\Leftrightarrow $ $\frac{\pi }{4}-x$ = $2x+k\pi $
$\Leftrightarrow $ $-2x-x = -\frac{\pi }{4}+k\pi $
$\Leftrightarrow $ $-3x=-\frac{\pi }{4}+k\pi $
$\Leftrightarrow $ $x=\frac{\pi }{12}-k\frac{\pi }{3}$ $\left ( k \epsilon Z \right )$
Cách 2:
$\tan \left ( \frac{\pi }{4}-x \right )$ = $\tan 2x$
$\Leftrightarrow $ $2x=\frac{\pi }{4}-x+k\pi $
$\Leftrightarrow $ $3x=\frac{\pi }{4}+k\pi $
$\Leftrightarrow $ $x=\frac{\pi }{12}+k\frac{\pi }{3} \left ( k\epsilon Z\right )$