ĐKXĐ: ...
Sử dụng công thức \(tana+tanb=\frac{sin\left(a+b\right)}{cosa.cosb}\) ta có:
\(tan\left(\frac{\pi}{3}-4x\right)+tan\left(\frac{\pi}{6}+x\right)=\frac{cosx}{sinx}-\frac{sin2x}{cos2x}\)
\(\Leftrightarrow\frac{sin\left(\frac{\pi}{2}-3x\right)}{cos\left(\frac{\pi}{3}-4x\right)cos\left(\frac{\pi}{6}+x\right)}=\frac{cos2x.cosx-sin2x.sinx}{sinx.cos2x}\)
\(\Leftrightarrow\frac{cos3x}{cos\left(\frac{\pi}{3}-4x\right)cos\left(\frac{\pi}{6}+x\right)}=\frac{cos3x}{sinx.cos2x}\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\Rightarrow...\\cos\left(\frac{\pi}{3}-4x\right)cos\left(\frac{\pi}{6}+x\right)=sinx.cos2x\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow cos\left(\frac{\pi}{2}-3x\right)+cos\left(\frac{\pi}{6}-5x\right)=sin3x-sinx\)
\(\Leftrightarrow sin3x+cos\left(\frac{\pi}{6}-5x\right)=sin3x-sinx\)
\(\Leftrightarrow cos\left(\frac{\pi}{6}-5x\right)=-sinx=cos\left(\frac{\pi}{2}+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{\pi}{6}-5x=\frac{\pi}{2}+x+k2\pi\\\frac{\pi}{6}-5x=-\frac{\pi}{2}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
