Question

If $5\cos 2\theta +2{\cos }^2\frac{\theta }{2}+1=0,-\pi <\theta <\pi$ then $\theta =$

• A. $\frac{\pi }{3}$
• B. ${\cos }^{-1}\frac{3}{5}$
• C. $-{\cos }^{-1}\frac{3}{5}$
• D. $\pi -{\cos }^{-1}\frac{3}{5}$

Answer 1

Answer: (A), (D)

$\frac{\pi }{3}$
$\pi -{\cos }^{-1}\frac{3}{5}$

Given, $5\cos 2\theta +1+\cos \theta +1=0$
$5(2{\cos }^2\theta -1)+\cos \theta +2=0$
$10{\cos }^2\theta +\cos \theta -3=0$
$\cos \theta =\frac{-1\pm 11}{20}=\frac{1}{2}$ or $\frac{-3}{5}$
$\therefore \theta =\frac{\pi }{3},\pi -{\cos }^{-1}\frac{3}{5}$