$a)\quad \tan(3x +1)=1$
$ĐKXĐ:\cos(3x +1)\ne 0 \Leftrightarrow x \ne \dfrac{\pi}{6}-\dfrac13 +\dfrac{n\pi}{3}$
$PT\Leftrightarrow \tan(3x +1)=\tan\dfrac{\pi}{4}$
$\Leftrightarrow 3x +1 =\dfrac{\pi}{4} + k\pi$
$\Leftrightarrow x =\dfrac{\pi}{12} -\dfrac13 +k\dfrac{\pi}{3}\quad (k\in\Bbb Z)$
$b)\quad \cos(2x +1)=\cos(2x -1)$
$\Leftrightarrow \left[\begin{array}{l}2x +1 = 2x - 1+k2\pi\\2x +1=1 - 2x + k2\pi\end{array}\right.$
$\Leftrightarrow 4x = k2\pi$
$\Leftrightarrow x =k\dfrac{\pi}{2}\quad (k\in\Bbb Z)$
