270 độ<x<360 độ
=>sinx<0 và cosx>0
\(cos2x=\dfrac{2}{3}\)
=>\(2\cdot cos^2x-1=\dfrac{2}{3}\)
=>\(2\cdot cos^2x=\dfrac{5}{3}\)
=>\(cos^2x=\dfrac{5}{6}\)
mà cosx>0
nên \(cosx=\dfrac{\sqrt{30}}{6}\)
=>\(sinx=-\dfrac{\sqrt{6}}{6}\)
\(sin\left(x-\dfrac{pi}{6}\right)=sinx\cdot cos\left(\dfrac{pi}{6}\right)-cosx\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=-\dfrac{\sqrt{6}}{6}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{30}}{6}\cdot\dfrac{1}{2}=\dfrac{-3\sqrt{2}-\sqrt{30}}{12}\)
\(cos\left(x-\dfrac{pi}{6}\right)=cosx\cdot cos\left(\dfrac{pi}{6}\right)+sinx\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{\sqrt{30}}{6}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{-\sqrt{6}}{6}\cdot\dfrac{1}{2}=\dfrac{\sqrt{90}-\sqrt{6}}{12}\)
