If the solution for theta of cos ptheta + co | vedclass

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Question

(b) Given $\cos p	heta = - \cos q	heta = \cos (\pi + q	heta )$

==> $p	heta = 2n\pi \pm (\pi + q	heta ),n \in I$

==> $	heta = \frac{{

Vedclass · Accepted Answer

$\frac{{2\pi }}{{p + q}}$

Vedclass · Answer

(b) Given $\cos p	heta = - \cos q	heta = \cos (\pi + q	heta )$

==> $p	heta = 2n\pi \pm (\pi + q	heta ),n \in I$

==> $	heta = \frac{{(2n + 1)\pi }}{{p - q}}$ or $\frac{{(2n - 1)\pi }}{{p + q}},\,\,n \in I$

Both the solutions form an $A.P.$ $	heta = \frac{{(2n + 1)\pi }}{{p - q}}$ gives us an $A.P.$ with common difference $\frac{{2\pi }}{{p - q}}$ and

$	heta = \frac{{(2n - 1)\pi }}{{p + q}}$ gives us an $A.P.$ with common difference $ = \frac{{2\pi }}{{p + q}}$. 

Certainly, $\frac{{2\pi }}{{p + q}} < \,\left| {\,\frac{{2\pi }}{{p - q}}\,} ight|$.

If the solution for $\theta $ of $\cos p\theta + \cos q\theta = 0,\;p > 0,\;q > 0$ are in $A.P.$, then the numerically smallest common difference of $A.P.$ is

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