\(\hept{\begin{cases}\cos x\ne0\\tan2x+1\ne0,cos2x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne\frac{\pi}{2}+k\pi\\x\ne-\frac{\pi}{8}+\frac{k\pi}{2}\\x\ne\frac{\pi}{4}+\frac{k\pi}{2}\end{cases}}}\)

\(1.TXĐ:sin4x\ne0\Leftrightarrow4x\ne k\pi\Leftrightarrow x\ne\frac{k\pi}{4}\left(k\in Z\right)\)

\(2,TXĐ:cos\left(x-2\pi\right)\ne0\Leftrightarrow x-2\pi\ne\frac{\pi}{2}+k\pi\Leftrightarrow x\ne\frac{5\pi}{2}+k\pi\left(k\in Z\right)\)

\(3,TXĐsin\left(\frac{3x}{2}+\frac{\pi}{4}\right)\ne0\Leftrightarrow\frac{3x}{2}+\frac{\pi}{4}\ne k\pi\Leftrightarrow....\)

mấy câu còn lại cx tương tự như vậy cho mẫu khác 0 còn câu 5 thì\(cos\left(\frac{x}{2}-\frac{\pi}{6}\right)\ne0\Leftrightarrow....\)

\(Pt\Leftrightarrow\orbr{\begin{cases}\frac{x}{2}+\frac{\pi}{5}=arcsin\left(\frac{-\sqrt{2}}{3}\right)+k2\pi\\\frac{x}{2}+\frac{\pi}{5}=\pi-arcsin\left(\frac{-\sqrt{2}}{3}\right)+k2\pi\end{cases}}\left(k\in Z\right)\Leftrightarrow....\)

\(\Leftrightarrow\frac{\sin^2x}{\cos^2x}-\frac{\cos^2x}{\sin^2x}=2\)

\(\Leftrightarrow\sin^4x-\cos^4x=2\sin^2x\cos^2x\)

\(\Leftrightarrow\left(\sin^2x-\cos^2x\right)\left(\sin^2x+\cos^2x\right)=2\sin^2x\cos^2x\)

\(\Leftrightarrow-2\left(\cos^2x-\sin^2x\right)=4\sin^2x\cos^2x\)

\(\Leftrightarrow-2\cos2x=\sin^22x=1-\cos^22x\)

\(\Leftrightarrow\cos^22x-2\cos2x-1=0\)

Giải PT bậc 2 tìm cosx từ đó suy ra x ban tự làm nốt nhé