If f (x) =$Co{s}^{2}x+Co{s}^{2}2x+Co{s}^{2}3x$ then the number of values of $x\in [0, 2\pi ]$ for which f(x) = 1 is

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\cos }^2\mathrm{x}+{\cos }^{2}2\mathrm{x}+{\cos }^{2}3\mathrm{x}; \mathrm{f}\left(\mathrm{x}\right)=1 \\ {\cos }^{2}3\mathrm{x}+{\cos }^{2}2\mathrm{x}+{\cos }^2\mathrm{x}=1 \\ 1+{\cos }^{2}3\mathrm{x}-{\sin }^{2}2\mathrm{x}+{\cos }^2\mathrm{x}=1 \\ \cos 5\mathrm{x}.\cos \mathrm{x}+{\cos }^2\mathrm{x}=0 \\ \cos \mathrm{x}(\cos 5\mathrm{x}+\cos \mathrm{x})=0 \\ \cos \mathrm{x}=0, \cos 5\mathrm{x}=-\cos \mathrm{x} \end{gathered}$$

$$\begin{gathered} \mathrm{x}=\frac{\mathrm{\pi }}{2}, \frac{3\mathrm{\pi }}{2} \cos 5\mathrm{x}=\cos (\mathrm{\pi }-\mathrm{x}) \\ 5\mathrm{x}=2\mathrm{n\pi }\pm \mathrm{\pi }-\mathrm{x} \\ 6\mathrm{x}=(2\mathrm{n}\pm 1) \mathrm{\pi } \Rightarrow \mathrm{x}=(2\mathrm{n}\pm 1) \frac{\mathrm{\pi }}{6} \\ \mathrm{x}=\frac{\mathrm{\pi }}{2}, 3\frac{\mathrm{\pi }}{6}, \frac{\mathrm{\pi }}{6}, \frac{\mathrm{\pi }}{2}, \frac{5\mathrm{\pi }}{6}, \frac{7\mathrm{\pi }}{6}, \frac{3\mathrm{\pi }}{2}, \frac{11\mathrm{\pi }}{6} \\ \mathrm{No}. \mathrm{of} \mathrm{solutions} 6 \end{gathered}$$