$1.b)cos(x-\dfrac{\pi}{4})=-\dfrac{1}{2}=cos(\dfrac{2\pi}{3})$
$⇔$\(\left[ \begin{array}{l}x-\dfrac{\pi}{4}=\dfrac{2\pi}{3}+k2\pi\\x-\dfrac{\pi}{4}=-\dfrac{2\pi}{3}+k2\pi\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{11\pi}{12}+k2\pi\\x=-\dfrac{5\pi}{12}+k2\pi\end{array} \right.\)$(k∈Z)$
$f)tan(\dfrac{x}{2}-\dfrac{\pi}{4})=tan\dfrac{\pi}{8}$
$⇔\dfrac{x}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{8}+k\pi$
$⇔\dfrac{x}{2}=\dfrac{3\pi}{8}+k\pi$
$⇔x=\dfrac{3\pi}{4}+k2\pi$$(k∈Z)$
$g)tan2x=cot\dfrac{2\pi}{7}=tan(\dfrac{\pi}{2}-\dfrac{2\pi}{7})$
$⇔2x=\dfrac{\pi}{2}-\dfrac{2\pi}{7}+k\pi$
$⇔x=\dfrac{3\pi}{28}+k2\pi$$(k∈Z)$
