$\begin{align}  & \text{From the L}\text{.H}\text{.S } \\ & 4\cos \theta \cos 3\theta +1=\frac{\sin \theta }{\sin \theta }\left[ 4\cos \theta \cos 3\theta +1 \right] \\ & 4\cos \theta \cos 3\theta +1=\frac{4\sin \theta \cos \theta \cos 3\theta +\sin \theta }{\sin \theta } \\ & 4\cos \theta \cos 3\theta +1=\frac{2\sin 2\theta \cos 3\theta +\sin \theta }{\sin \theta } \\ & \text{Convert }\sin 2\theta \cos 3\theta \text{ to sum of sine function} \\ & \text{Note: }\sin \theta \cos \beta =\frac{1}{2}\left[ \sin (\theta +\beta )+\sin (\theta -\beta ) \right] \\ & 4\cos \theta \cos 3\theta +1=\frac{\sin 5\theta -\sin \theta +\sin \theta }{\sin \theta } \\ & 4\cos \theta \cos 3\theta +1=\frac{\sin 5\theta }{\sin \theta }----(R.H.S) \\ & \cos 3\theta =\cos (2\theta +\theta )=\cos 2\theta \cos \theta -\sin 2\theta \sin \theta  \\ & \cos 3\theta =(2{{\cos }^{2}}\theta -1)\cos \theta -(2\sin \theta \cos \theta )\sin \theta  \\ & \cos 3\theta =(2{{\cos }^{3}}\theta -\cos \theta )-2{{\sin }^{2}}\theta \cos \theta  \\ & \cos 3\theta =2{{\cos }^{3}}\theta -\cos \theta -2(1-{{\cos }^{2}}\theta )\cos \theta  \\ & 3\cos 3\theta =2{{\cos }^{3}}\theta -\cos \theta -2\cos \theta +2{{\cos }^{3}}\theta  \\ & 3\cos 3\theta =4{{\cos }^{3}}\theta -3\cos \theta ------(i) \\ & \cos \theta \cos 3\theta =\frac{1}{2} \\ & \cos \theta \cos 3\theta =\cos \theta (4{{\cos }^{3}}\theta -3\cos \theta )=\frac{1}{2} \\ & 4{{\cos }^{4}}\theta -3{{\cos }^{2}}\theta =\frac{1}{2} \\ & 8{{\cos }^{4}}\theta -6{{\cos }^{2}}\theta -1=0 \\ & \text{Let }x={{\cos }^{2}}\theta  \\ & 8{{x}^{2}}-6x-1=0 \\ & \text{Using quadratic formula} \\ & x=\frac{6\pm \sqrt{36+32}}{16}=\frac{6\pm \sqrt{68}}{16} \\ & x=0.8903\text{ or }x=-0.1404 \\ & \text{But }x={{\cos }^{2}}\theta  \\ & \cos \theta =\sqrt{x} \\ & \text{when }x=0.8903 \\ & x=\sqrt{0.8903}=\pm 0.9436 \\ & \theta =\cos \theta (\pm 0.9436)={{19.33}^{\circ }},{{160.67}^{\circ }} \\ & \text{when }x=-0.1404,\text{ there is no solution} \\\end{align}$