$tan\frac{B}2.tan\frac{C}2 = \frac{1}3 \Leftrightarrow 3sin\frac{B}2.sin\frac{C}2 = cos\frac{B}2.cos\frac{C}2$
$\Leftrightarrow sin\frac{B}2.sin\frac{C}2 = cos\frac{B}2.cos\frac{C}2 - 2sin\frac{B}2.sin\frac{C}2$
$\Leftrightarrow cos\frac{B}2.cos\frac{C}2 + sin\frac{B}2.sin\frac{C}2 = 2cos\frac{B}2.cos\frac{C}2 - 2sin\frac{B}2.sin\frac{C}2$
$\Leftrightarrow cos(\frac{B}2 - \frac{C}2) = 2cos(\frac{B}2+\frac{C}2)$
Thế vào PT có
$x^2+x+cos(B+C)-\frac1{4}cos(B-C)=0$
$\Leftrightarrow x^2+x+2cos^2(\frac{B+C}2)-1-\frac{1}4[2cos^2(\frac{B-C}2)-1]=0$
$\Leftrightarrow x^2+x+2cos^2(\frac{B+C}2)-1-\frac{1}2.[4cos^2(\frac{B+C}2)]+\frac{1}4=0$
$\Leftrightarrow x^2+x-1+\frac{1}4=0$
$\Leftrightarrow x^2+x-\frac{3}4=0$
$\Leftrightarrow x=\frac{1}2;-\frac{3}2$