根据题意有$$ \left(\cos\alpha+\sin\alpha\right)^{2}=\dfrac{1}{9},$$所以$$\sin\alpha\cos\alpha=-\dfrac{4}{9},$$结合 $\alpha\in\left(0,{\mathrm \pi} \right)$ 可知$$\sin\alpha>0,\cos\alpha<0.$$所以$$\cos\alpha-\sin\alpha=-\sqrt{\left(\cos\alpha+\sin\alpha\right)^{2}-4\sin\alpha\cos\alpha}=-\dfrac{\sqrt{17}}{3},$$进而可得$$\cos2\alpha=\left(\cos\alpha+\sin\alpha\right)\left(\cos\alpha-\sin\alpha\right)=\dfrac{\sqrt{17}}{9}.$$