The ratio of the areas bounded by y = cos x, y = cos 2x between x = 0 , $\mathrm{x}=\frac{\mathrm{\pi }}{3}$ and the X-axis is

$A_1={\int }_0^{\pi /3} \cos xdx=[\sin x{]}_0^{\pi /3}=\frac{\sqrt{3}}{2}$

$y=\cos 2x\Rightarrow \cos 2x=0\Rightarrow 2x=-\frac{\pi }{2},\frac{\pi }{2}$

$\Rightarrow x=-\pi /4,\pi /4$

$A_2={\int }_0^{\pi /4} \cos 2xdx-{\int }_{\pi /4}^{\pi /3} \cos 2xdx$

$={\left[\frac{\sin 2x}{2}\right]}_0^{\pi /4}-{\left[\frac{\sin 2x}{2}\right]}_{\pi /4}^{\pi /3}=\frac{1}{2}\left[1\right]-\frac{1}{2}\left[\frac{\sqrt{3}}{2}-1\right]=1-\frac{\sqrt{3}}{4}$

$=\frac{\sqrt{3}}{2}:1-\frac{\sqrt{3}}{4}=2\sqrt{3}:4-\sqrt{3}.$