Question

\(2\left(\sqrt{3}sinx+cosx\right)=3sin2x+\sqrt{7}cos2x\)

Answer 1

\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx=\frac{3}{4}sin2x+\frac{\sqrt{7}}{4}cos2x\)

Đặt \(cosa=\frac{3}{4}\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=sin\left(2x+a\right)\)

\(\Rightarrow\left[{}\begin{matrix}2x+a=x+\frac{\pi}{6}+k2\pi\\2x+a=\frac{5\pi}{6}-x+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}-a+k2\pi\\x=-\frac{a}{3}+\frac{5\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)