\(sin^2x+cos^2x=1\)
=>\(sin^2x=1-\dfrac{9}{16}=\dfrac{7}{16}\)
=>\(\left[{}\begin{matrix}sinx=\dfrac{\sqrt{7}}{4}\\sinx=-\dfrac{\sqrt{7}}{4}\end{matrix}\right.\)
\(A=sin\left(x+\dfrac{\Omega}{3}\right)=sinx\cdot cos\left(\dfrac{\Omega}{3}\right)+cosx\cdot sin\left(\dfrac{\Omega}{3}\right)\)
\(=\dfrac{1}{2}\cdot sinx+cosx\cdot\dfrac{\sqrt{3}}{2}\)
\(=\dfrac{1}{2}\cdot sinx+\dfrac{-3\sqrt{3}}{8}\)
TH1: \(sinx=\dfrac{\sqrt{7}}{4}\)
=>\(A=\dfrac{1}{2}\cdot\dfrac{\sqrt{7}}{4}-\dfrac{3\sqrt{3}}{8}=\dfrac{\sqrt{7}-3\sqrt{3}}{8}\)
TH2: \(sinx=-\dfrac{\sqrt{7}}{4}\)
=>\(A=\dfrac{-1}{2}\cdot\dfrac{\sqrt{7}}{4}-\dfrac{3\sqrt{3}}{8}=\dfrac{-\sqrt{7}-3\sqrt{3}}{8}\)
\(B=sin\left(x-\dfrac{\Omega}{3}\right)=sinx\cdot cos\left(\dfrac{\Omega}{3}\right)-cosx\cdot sin\left(\dfrac{\Omega}{3}\right)\)
\(=sinx\cdot\dfrac{1}{2}-cosx\cdot\dfrac{\sqrt{3}}{2}\)
\(=\dfrac{1}{2}\cdot sinx+\dfrac{3\sqrt{3}}{8}\)
TH1: \(sinx=-\dfrac{\sqrt{7}}{4}\)
=>\(B=\dfrac{1}{2}\cdot\dfrac{-\sqrt{7}}{4}+\dfrac{3\sqrt{3}}{8}=\dfrac{3\sqrt{3}-\sqrt{7}}{8}\)
TH2: \(sinx=\dfrac{\sqrt{7}}{4}\)
=>\(B=\dfrac{1}{2}\cdot\dfrac{\sqrt{7}}{4}+\dfrac{3\sqrt{3}}{8}=\dfrac{3\sqrt{3}+\sqrt{7}}{8}\)
