Trigonometric Sequence

By expanding $\cos(2a)$ and $\cos(3a)$ into $2\cos(a)^2-1$ and $4\cos(a)^3-3\cos(a)$ , and using the fact that $\cos(a)+\cos(3a)=2\cos(2a)$ since the three terms are in an arithmetic progression, the equation can be solved for $a$ . Substitution and simplification leads to $4\cos(a)^3-4\cos(a)^2-2\cos(a)+2=0$ , which can be factored into $(4\cos(a)^2-2)(\cos(a)-1)=0$ . This means that $\cos(a)^2=0.5$ and $\cos(a)=1$ . When $\cos(a)=1$ , $a=0°$ and $a=360°$ which are excluded from the domain $0°<a<360°$ . When $\cos(a)^2=0.5$ , then $a=45°, 135°, 225°$ , and $315°$ . Summing these values leads to a final answer of $720°$ .

given cos(a), cos(2a), cos(3a) are in AP then 2cos(2a)=cos(a) +cos(3a) which gives cos(2a) = cos(2a)*cos(a) by cosC +cosD formula from here we get cos(a)=1 which gives a=0 and a=360 but it's not in our domain and cos(2a)=0 which gives a=45 ,a=135, a= 225 ,a= 315 and sum is 720