If the solutions of $\cos p\theta +\cos q\theta =0$  are in A.P., then the common difference of A.P. is

$2.\cos \left(\frac{p+q}{2}\right)\theta .\cos \left(\frac{p-q}{2}\right)\theta =0$

$\cos \left(\frac{p+q}{2}\right)\theta =0\left(or\right)\cos \left(\frac{p-q}{2}\right)\theta =0$

$\therefore \left(\frac{p+q}{2}\right)\theta =\left(2n+1\right)\frac{\pi }{2}\left(or\right)\left(\frac{p-q}{2}\right)\theta =\left(2n+1\right)\frac{\pi }{2}$

When $\therefore \theta =\frac{\left(2n+1\right)\pi }{p+q}$

Sequence is $\frac{3\pi }{p+q},\frac{5\pi }{p+q},\frac{7\pi }{p+q}\mathrm{...}$   common difference is $\frac{2\pi }{p+q}$

When $\therefore \theta =\frac{\left(2n+1\right)\pi }{p-q}$

Sequence is $\frac{3\pi }{p-q},\frac{5\pi }{p-q},\frac{7\pi }{p-q}\mathrm{...}$   common difference is $\frac{2\pi }{p-q}$