<s>[INST] Acute triangle $\triangle ABC$ has $(\sin A - 2 \sin 2B) = 2 - 2 \cos 2B$ Find the range of $\frac{\sin B + \sin C}{\sin A}$ Angles in acute triangle $\triangle ABC$ has $(\sin A - 2 \sin 2B) \tan A = 2 - 2 \cos 2B$
Find the range of $$\frac{\sin B + \sin C}{\sin A}$$
We can prove that $\frac{a^2}{bc} = 4$ through some tedious trigonometric computations
Let $\frac{b+c}{a} = k$, then
$1 > \cos A = \frac{b^2+c^2-a^2}{2bc}$, and we plug in  $\frac{a^2}{bc} = 4$, and we might be able to find the range of $k$ somehow? But I can't finish and the computation seems daunting. I am also not sure if $\cos A <1$ condition is too loose.
 [/INST] Looks like you are very close.
You have established $\frac {\sin B + \sin C} {\sin A} = \frac {b+c}a$ using the Law of Sines.
From $\frac {a^2}{bc} = 4$ we obtain $a^2 = 4bc$ and $bc = \frac{a^2}4$.
Now:
\begin{align}
1 &> \cos A\\ &= \frac {b^2 + c^2 - a^2}{2bc}\\
&=\frac {b^2 + c^2 - 4bc}{2bc}\\
&=\frac {b^2 + 2bc + c^2 - 6bc}{2bc}\\
&=\frac {(b+c)^2}{a^2/2}-6\\
&=2k^2-3\\
k^2& < 2
\end{align}
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