- Home
- Standard 11
- Mathematics
Trigonometrical Equations
medium
જો $\sqrt 3 \cos \,\theta + \sin \theta = \sqrt 2 ,$ તો $\theta $ નો વ્યાપક ઉકેલ મેળવો.
A
$n\pi + {( - 1)^n}\frac{\pi }{4}$
B
${( - 1)^n}\frac{\pi }{4} - \frac{\pi }{3}$
C
$n\pi + \frac{\pi }{4} - \frac{\pi }{3}$
D
$n\pi + {( - 1)^n}\frac{\pi }{4} - \frac{\pi }{3}$
Solution
(d) $\frac{{\sqrt 3 }}{2}\cos \theta + \frac{1}{2}\sin \theta = \frac{{\sqrt 2 }}{2}$ { dividing by $\sqrt {{{(\sqrt 3 )}^2} + {1^2}} = 2 $}
$ \Rightarrow $ $\sin \left( {\theta + \frac{\pi }{3}} \right) = \frac{1}{{\sqrt 2 }} = \sin \left( {\frac{\pi }{4}} \right)$
$ \Rightarrow $ $\theta = n\pi + {( – 1)^n}\frac{\pi }{4} – \frac{\pi }{3}$.
Standard 11
Mathematics
