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સમીકરણ $2\sqrt 3 \cos \theta = \tan \theta $ નું સમાધાન કરે તેવા $\theta $ નો વ્યાપક ઉકેલ મેળવો.
A
$2n\pi \pm \frac{\pi }{6}$
B
$2n\pi \pm \frac{\pi }{4}$
C
$n\pi + {( - 1)^n}\frac{\pi }{3}$
D
$n\pi + {( - 1)^n}\frac{\pi }{4}$
Solution
(c) $2\sqrt 3 {\cos ^2}\theta = \sin \theta $
==> $2\sqrt 3 {\sin ^2}\theta + \sin \theta – 2\sqrt 3 = 0$
==> $\sin \theta = \frac{{ – 1 \pm 7}}{{4\sqrt 3 }}$
$\Rightarrow \sin \theta = \frac{{ – 8}}{{4\sqrt 3 }}{\rm{,}}$ (Impossible)
and $\sin \theta = \frac{6}{{4\sqrt 3 }} = \frac{{\sqrt 3 }}{2}$
==> $\theta = n\pi + {( – 1)^n}\frac{\pi }{3}$.
Standard 11
Mathematics
