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Trigonometrical Equations
hard
જો $\cos 2\theta = (\sqrt 2 + 1)\,\,\left( {\cos \theta - \frac{1}{{\sqrt 2 }}} \right)$, તો $\theta $ નો વ્યાપક ઉકેલ મેળવો.
A
$2n\pi + \frac{\pi }{4}$
B
$2n\pi \pm \frac{\pi }{4}$
C
$2n\pi - \frac{\pi }{4}$
D
એકપણ નહિ.
Solution
(b) $2{\cos ^2}\theta – (\sqrt 2 + 1)\cos \theta – 1 + \frac{{(\sqrt 2 + 1)}}{{\sqrt 2 }} = 0$
$ \Rightarrow $ $\cos \theta = \frac{{(\sqrt 2 + 1) \pm \sqrt {{{(\sqrt 2 + 1)}^2} – \frac{8}{{\sqrt 2 }}} }}{4}$
$ \Rightarrow $ $\cos \theta = \cos \left( {\frac{\pi }{4}} \right)$
$ \Rightarrow $ $\theta = 2n\pi \pm \frac{\pi }{4}$.
Trick : Since $\theta = \frac{\pi }{4}$ satisfies the equation and therefore the general value should be $2n\pi \pm \frac{\pi }{4}$.
Standard 11
Mathematics
