Question

\(tanx-3cotx=6\) \(\left(\pi< x< \frac{3\pi}{2}\right)\)

Tính

a)\(sinx+cosx\)

b)\(\frac{2sinx-tanx}{cosx+cotx}\)

Answer 1

\(\pi< x< \frac{3\pi}{2}\Rightarrow sinx< 0;cosx< 0;tanx>0;cotx>0\)

\(tanx-3cotx=6\Leftrightarrow tanx-\frac{3}{tanx}=6\)

\(\Leftrightarrow tan^2x-6tanx-3=0\Rightarrow\left[{}\begin{matrix}tanx=3+2\sqrt{3}\\tanx=3-2\sqrt{3}< 0\left(l\right)\end{matrix}\right.\)

\(\frac{1}{cos^2x}=1+tan^2x\Rightarrow cos^2x=\frac{1}{1+tan^2x}\Rightarrow cosx=\frac{-1}{\sqrt{1+tan^2x}}\) (do \(cosx< 0\))

\(\Rightarrow cosx=\frac{-1}{\sqrt{22+12\sqrt{3}}}\Rightarrow sinx=-\sqrt{1-cos^2x}=-\sqrt{\frac{15+6\sqrt{3}}{26}}\)

\(cotx=\frac{1}{tanx}=\frac{1}{3+2\sqrt{3}}\)

Số xấu dữ dội, bạn tự thay vào kết quả :(