Trigonometrical Equations
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જો સમીકરણ $\cos p\theta + \cos q\theta = 0,\;p > 0,\;q > 0$ ની $\theta $ ના ઉકેલગણ સમાંતર શ્રેણીમાં હોય તો સમાંતર શ્રેણીનો ન્યુનતમ સમાન્ય તફાવત મેળવો.

A

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

B

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

C

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

D

$\frac{1}{{p + q}}$

Solution

(b) Given $\cos p\theta = – \cos q\theta = \cos (\pi + q\theta )$

==> $p\theta = 2n\pi \pm (\pi + q\theta ),n \in I$

==> $\theta = \frac{{(2n + 1)\pi }}{{p – q}}$ or $\frac{{(2n – 1)\pi }}{{p + q}},\,\,n \in I$

Both the solutions form an $A.P.$ $\theta = \frac{{(2n + 1)\pi }}{{p – q}}$ gives us an $A.P.$ with common difference $\frac{{2\pi }}{{p – q}}$ and

$\theta = \frac{{(2n – 1)\pi }}{{p + q}}$ gives us an $A.P.$ with common difference $ = \frac{{2\pi }}{{p + q}}$.

Certainly, $\frac{{2\pi }}{{p + q}} < \,\left| {\,\frac{{2\pi }}{{p – q}}\,} \right|$.

Standard 11
Mathematics

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