Question

If $\cos p\theta +cosq\theta =0$ then the different values $\theta$ are in A.P with common difference $2\pi (p+q)$.

Answer 1

$\cos p\theta +\cos q\theta =0\Rightarrow 2\cos \left(\frac{p+q}{2}\right)\theta \cos \left(\frac{p-q}{2}\right)\theta =0$
$\Rightarrow \cos \left(\frac{p+q}{2}\right)\theta =0$ or $\cos \left(\frac{p-q}{2}\right)\theta =0$
$\Rightarrow \left(\frac{p+q}{2}\right)\theta =(2n+1)\frac{\pi }{2}$ or $\left(\frac{p-q}{2}\right)\theta =(2n+1)\frac{\pi }{2}$
$\Rightarrow \theta =\frac{(2n+1)\pi }{p+q}$ or $\theta =\frac{(2n+1)\pi }{p-q}$
Hence $\theta =\frac{\pi }{p+q},\frac{3\pi }{p+q}...\frac{\pi }{p-q},\frac{3\pi }{p-q}...\to A.P$