If cospθ+cosqθ=0, prove that the different values of θ form two arithmetical progressions in which the common difference are

2∏/(p+q) and 2∏/(p-q) resectively

we get 

cos px=-cos qx

cos px=cos (pi-qx)

px=n(pi)+/-qx

taking the first case

px=n(pi)+qx

x(p-q)=n(pie)

x=n(??pie)/p-q

taking the second case we get

x=n(??pie)/p+q;

we find the common difference is what is asked

note

pie denotes the constat 3.14....