The number of solutions of the trigonometric equations $1+\cos x.\cos 5x={\sin }^{2}x$ in $\left[0,2\pi \right]$ is

$\cos x.\cos 5x+{\cos }^{2}x=0$ 

$\Rightarrow \cos x=0$ (or) $\cos 5x+\cos x=0$

$\Rightarrow \cos x=0, 2\cos 3x.\cos 2x=0$

$\Rightarrow x=(2n+1)\frac{\pi }{2}$, $3x=(2n+1)\frac{\pi }{2}, 2x=(2n+1)\frac{\pi }{2}$

$x=(2n+1)\frac{\pi }{2}, x=(2n+1)\frac{\pi }{6}, x=(2n+1)\frac{\pi }{4}$

$\therefore Set of solutions is \{\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6},\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}\}$ $\left(\because x\in \left[0,2\mathrm{\pi }\right]\right)$