Solve $\sin\left(x-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$sin(x−π4)=1√2 for $-\pi\le x$−π≤x$<$<$\pi$π.
Solve $\tan\left(2x-\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}}$tan(2x−π6)=−1√3 for $0\le x\le\pi$0≤x≤π.