- Home
- Standard 11
- Mathematics
Trigonometrical Equations
hard
यदि $5\cos 2\theta + 2{\cos ^2}\frac{\theta }{2} + 1 = 0, - \pi < \theta < \pi $, तब $\theta = $
A
$\frac{\pi }{3}$
B
$\frac{\pi }{3},{\cos ^{ - 1}}\frac{3}{5}$
C
${\cos ^{ - 1}}\frac{3}{5}$
D
$\frac{\pi }{3},\pi - {\cos ^{ - 1}}\frac{3}{5}$
Solution
$5\cos 2\theta + 2{\cos ^2}\frac{\theta }{2} + 1 = 0$
$ \Rightarrow $ $5(2{\cos ^2}\theta – 1) + (1 + \cos \theta ) + 1 = 0$
$ \Rightarrow $ $10{\cos ^2}\theta + \cos \theta – 3 = 0$
$ \Rightarrow $ $(5\cos \theta + 3)\,(2\cos \theta – 1) = 0$
$ \Rightarrow $ $\cos \theta = \frac{1}{2},\,\cos \theta = – \frac{3}{5}$
$\Rightarrow \theta = \frac{\pi }{3},\,\pi – {\cos ^{ – 1}}\left( {\frac{3}{5}} \right)$.
Standard 11
Mathematics
