Ta có \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-\dfrac{1}{9}=\dfrac{8}{9}\Leftrightarrow cosx=\pm\dfrac{2\sqrt{2}}{3}\)
Với \(cosx=\dfrac{2\sqrt{2}}{3}\)
\(cos2x=1-2sin^2x=1-\dfrac{2.1}{9}=1-\dfrac{2}{9}=\dfrac{7}{9}\)
\(sin2x=2sinx.cosx=\dfrac{2.1}{3}.\dfrac{2\sqrt{2}}{3}=\dfrac{4\sqrt{2}}{9}\)
\(cos\left(150^0+x\right)=cos150^0.cosx-sin150^0.sinx\)
\(=-\dfrac{\sqrt{3}}{2}.\dfrac{2\sqrt{2}}{3}-\dfrac{1}{2}.\dfrac{1}{3}=\dfrac{-\sqrt{6}}{3}-\dfrac{1}{6}=\dfrac{-2\sqrt{6}-1}{6}\)
Với \(cos=-\dfrac{2\sqrt{2}}{3}\)
\(sin2x=\dfrac{2}{3}.\left(-\dfrac{2\sqrt{2}}{3}\right)=\dfrac{-4\sqrt{2}}{9}\)
\(cos\left(150^0+x\right)=-\dfrac{\sqrt{3}}{2}.\left(-\dfrac{2\sqrt{2}}{3}\right)-\dfrac{1}{6}=\dfrac{\sqrt{6}}{3}-\dfrac{1}{6}=\dfrac{2\sqrt{6}-1}{6}\)
