4. In $\triangle A B C$, the sides opposite to angles $A, B, C$ are denoted as $a, b, c (b \neq 1)$, respectively, and $\frac{C}{A}, \frac{\sin B}{\sin A}$ are both roots of the equation $\log _{\sqrt{6}} x=$ $\log _{b}(4 x-4)$. Then $\triangle A B C$
(A) is an isosceles triangle, but not a right triangle
(B) is a right triangle, but not an isosceles triangle
(C) is an isosceles right triangle
(D) is neither an isosceles triangle nor a right triangle
Answer ( )
(B)
4.【Analysis and Solution】From the given equation, we get $\log _{b} x^{2}=\log _{b}(4 x-4)$, which means $x^{2}-4 x+4=0$,
The roots obtained are $x_{1}=x_{2}=2$.
Thus, $C=2A$ and $\sin B=2 \sin A$,
Since $A+B+C=180^{\circ}$, it follows that $3A+B=180^{\circ}$.
Therefore, $\sin B=\sin 3A$.
So, $2 \sin A-4 \sin ^{3} A=2 \sin A$,
which simplifies to $\sin A\left(1-4 \sin ^{2} A\right)=0$, but $\sin A \neq 0$,
Thus, $\sin ^{2} A=\frac{1}{4}$.
Since $\sin A>0$, we have $\sin A=\frac{1}{2}$,

Therefore, $A=30^{\circ}, C=60^{\circ}, B=90^{\circ}$.
Hence, the answer is (B).