Normal solution: 1. use the double angle formula to decompose, and recall cos^2(x)+sin^2(x)=1 cos(4x)=cos^2(2x)-sin^2(2x)=2cos^2(2x)-1 .................(1) 2. substitute (1) in (0) 2cos^2(2x)-1-cos(2x)=0 3. substitute u=cos(2x) 2u^2-u-1=0 4. Solve for x factor (u-1)(u+1/2)=0 => u=1 or u=-1/2 However, since cos(x) is an even function, so solutions to {cos(2x)=1, cos(-2x)=1, cos(2x)=-1/2 and cos(-2x)} ...........(2) are all solutions. 5. The cosine function is symmetrical about pi, therefore cos(-2x)=cos(2*pi-2x), solution (2) above becomes {cos(2x)=1, cos(2pi-2x)=1, cos(2x)=-1/2, cos(2pi-2x)=-1/2} 6. Solve each case cos(2x)=1 => x=0 cos(2pi-2x)=1 => cos(2pi-0)=1 => x=pi cos(2x)=-1/2 => 2x=2pi/3 or 2x=4pi/3 => x=pi/3 or 2pi/3 cos(2pi-2x)=-1/2 => 2pi-2x=2pi/3 or 2pi-2x=4pi/3 => x=2pi/3 or x=4pi/3 Summing up, x={0,pi/3, 2pi/3, pi, 4pi/3}
