Answer:
[tex]x=0^\circ,120^\circ,240^\circ,360^\circ[/tex]
Step-by-step explanation:
Use identities to set the equation up as a quadratic
[tex]\cos x=\cos 2x;\:0^\circ \leq x \leq 360^\circ\\\\\cos x=2\cos^2 x-1\\\\0=2\cos^2 x-\cos x-1[/tex]
Make the substitution u=cos(x) and solve the quadratic
[tex]0=2u^2-u-1\\\\0=(2u+1)(u-1)[/tex]
[tex]\displaystyle 0=2u+1\\\\0=2\cos x+1\\\\-1=2\cos x\\\\-\frac{1}{2}=\cos x\\\\120^\circ,240^\circ=x[/tex]
[tex]0=u-1\\\\0=\cos x-1\\\\1=\cos x\\\\0^\circ,360^\circ=x[/tex]
Hence, [tex]x=0^\circ,120^\circ,240^\circ,360^\circ[/tex]
