Question

Solve:

$5cos2\theta +2co{s}^2\frac{\theta }{2}+1=0,-\pi <\theta <\pi$

Answer 1

$5\cos 2\theta +2{\cos }^2\frac{\theta }{2}+1=0$

$5\cos 2\theta +2{\cos }^2\frac{\theta }{2}-1+1+1=0$ by adding $-1$ and $1$

$\Rightarrow 5\cos 2\theta +\cos \theta +2=0$

$\Rightarrow 5(2{\cos }^2\theta -1)+\cos \theta +2=0$ since $\cos 2\theta =2{\cos }^2\theta -1$

$\Rightarrow 10{\cos }^2\theta -5+\cos \theta +2=0$

$\Rightarrow 10{\cos }^2\theta +\cos \theta -3=0$

$\Rightarrow 10{\cos }^2\theta -5\cos \theta +6\cos \theta -3=0$

$\Rightarrow 5\cos \theta (2\cos \theta -1)+3(2\cos \theta -1)=0$

$\Rightarrow (2\cos \theta -1)(5\cos \theta +3)=0$

$\Rightarrow \cos \theta =\frac{1}{2},-\frac{3}{5}$

$\therefore \theta =\frac{\pi }{3},{\cos }^{-1}(-\frac{3}{5})$

$\therefore \theta =\{\frac{\pi }{3},-\pi +{\cos }^{-1}\frac{3}{5}\}$