Given the following question:
[tex]\begin{gathered} 8\cos ^2x+4\cos x=0 \\ (0,2\pi) \\ \end{gathered}[/tex][tex]\begin{gathered} 8\cos ^2x+4\cos x=0,(0,2\pi) \\ =8\cos ^2\mleft(x\mright)+4\cos \mleft(x\mright)=0,\: \mleft(0,\: 2\pi\mright) \\ =8\cos ^2\mleft(x\mright)+4\cos \mleft(x\mright)=0,\: \mleft(0,\: 2\pi\mright)=\cos \mleft(x\mright)=0,\: \cos \mleft(x\mright)=-\frac{1}{2}(\text{ By substitution)} \\ \cos (x)=0,\: \cos (x)=-\frac{1}{2} \\ \cos (x)=0,(0,2\pi)=x=\frac{\pi}{2}x=\frac{3\pi}{2} \\ \cos (x)=-\frac{1}{2},(0,2\pi)=x=\frac{2\pi}{3}x=\frac{4\pi}{3} \\ \text{ Combining all the solutions to get:} \\ x=\frac{\pi}{2} \\ x=\frac{3\pi}{2} \\ x=\frac{2\pi}{3} \\ x=\frac{4\pi}{3} \\ \text{ or options C, D, H, F} \end{gathered}[/tex]Your answers are options C, D, H, F.