you know i copy the net :))

HT

tanx=tan3pi/11

x=3pi/11+kpi 

\(\frac{\pi}{4}< \frac{3\pi}{11}+k\pi< 2\pi\)

\(\frac{1}{4}< \frac{3}{11}+k< 2\)   

\(\frac{1}{4}-\frac{3}{11}< k< 2-\frac{3}{11}\)   

\(-\frac{1}{44}< k< \frac{19}{11}\)   

\(\Rightarrow k=0;k=1\)   

Vậy chọn B 

\(cos^2x-sin^2x-2\sqrt{3}sinxcosx=1\)

\(\Leftrightarrow cos2x-\sqrt{3}sin2x=1\)

\(\Leftrightarrow\frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2x=\frac{1}{2}\)

\(\Leftrightarrow cos\left(\frac{\pi}{3}\right)cos2x-sin\left(\frac{\pi}{3}\right)sin2x=\frac{1}{2}\)

\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}\right)\)

\(\Leftrightarrow2x+\frac{\pi}{3}=\pm\frac{\pi}{3}+k2\pi\left(k\inℤ\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x=k\pi\\x=\frac{-\pi}{3}+k\pi\end{cases}}\left(k\inℤ\right)\)

\(cos^2x-sin^2x-2\sqrt{3}sinx.cosx=1\)

\(\Leftrightarrow cos2x-\sqrt{3}sin2x=1\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{3}\right)\Leftrightarrow2x+\frac{\pi}{3}=\pm\frac{\pi}{3}+k2\pi\)

\(\Leftrightarrow\orbr{\begin{cases}x=k\pi\\x=-\frac{\pi}{3}+k\pi\end{cases}}\)

ta có \(x\in\left[-\frac{\pi}{4};0\right]\Rightarrow2x\in\left[-\frac{\pi}{2},0\right]\Rightarrow sin2x\in\left[-1,0\right]\)

Vậy \(\hept{\begin{cases}GTNN=-1\\GTLN=0\end{cases}}\)

câu a là = căn2/2 nha mng

với cả cả a lẫn b đều là mũ 3 ạ