Sửa lại đề \(\dfrac{sinB}{sinC}=\dfrac{2cosC+cosA}{2cosB+cosA}\)
\(\Leftrightarrow sinB\left(2cosB+cosA\right)=sinC\left(2cosC+cosA\right)\)
\(\Leftrightarrow2sinBcosB+cosAsinB=2sinCcosC+cosAsinC\)
\(\Leftrightarrow cosA\left(sinB-sinC\right)+sin2B-sin2C=0\)
\(\Leftrightarrow2cosA.cos\dfrac{B+C}{2}.sin\dfrac{B-C}{2}+2cos\left(B+C\right)sin\left(B-C\right)=0\)
\(\Leftrightarrow2cosA.sin\dfrac{A}{2}.sin\dfrac{B-C}{2}-2cosA.2.sin\dfrac{B-C}{2}cos\dfrac{B-C}{2}=0\)
\(\Leftrightarrow2cosA.sin\dfrac{B-C}{2}\left(sin\dfrac{A}{2}-2cos\dfrac{B-C}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosA=0\\sin\dfrac{B-C}{2}=0\end{matrix}\right.\) \(\left(sin\dfrac{A}{2}-2cos\dfrac{B-C}{2}=\dfrac{\sqrt{2}}{2}-2.1\ne0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}A=\dfrac{\pi}{2}\\\dfrac{B-C}{2}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}A=\dfrac{\pi}{2}\\B=C\end{matrix}\right.\)
Khi \(\Delta ABC\) là \(\Delta\) vuông tại \(A\) hay là \(\Delta\) cân tại \(A\) thì thỏa đề bài.
