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

$$\begin{gathered} \cos \mathrm{p\theta }+\cos \mathrm{q\theta }=0 \\ \cos \mathrm{p\theta }=-\cos \mathrm{q\theta } \\ \cos \mathrm{p\theta }=\cos \left(\mathrm{\pi }-\mathrm{q\theta }\right) \\ \mathrm{p\theta }=2\mathrm{n\pi }\pm \left(\mathrm{\pi }-\mathrm{q\theta }\right) \\ (\mathrm{p}+\mathrm{q})\mathrm{\theta }=(2\mathrm{n}+1)\mathrm{\pi } \\ \mathrm{p\theta }=2\mathrm{n\pi }-\mathrm{\pi }+\mathrm{q\theta } \\ (\mathrm{p}-\mathrm{q})\mathrm{\theta }=(2\mathrm{n}-1)\mathrm{\pi } \\ \mathrm{\theta }=\frac{(2\mathrm{n}+1)\mathrm{\pi }}{\mathrm{p}+\mathrm{q}}, \frac{(2\mathrm{n}-1)\mathrm{\pi }}{\mathrm{p}-\mathrm{q}} \\ \mathrm{common} \mathrm{difference} \mathrm{of} \mathrm{A}.\mathrm{P} \\ \frac{2\mathrm{\pi }}{\mathrm{p}+\mathrm{q}}, \frac{2\mathrm{\pi }}{\mathrm{p}-\mathrm{q}} \end{gathered}$$