Question

Find the general solution of the given equation
$5\cos 2\theta +2{\cos }^2\frac{\theta }{2}+1=0$

Answer 1

Given trigonometric equation as:
$5\cos 2\theta +2{\cos }^2\frac{\theta }{2}+1=0$

To make the arguments uniform, apply
$\cos 2\theta =2{\cos }^2\theta -1$ and
$\cos \theta =2{\cos }^2\frac{\theta }{2}-1$

The original equation transforms to
$\Rightarrow 5(2{\cos }^2\theta -1)+(\cos \theta +1)+1=0$
$\Rightarrow 10{\cos }^2\theta -5+\cos \theta +2=0$
$\Rightarrow 10{\cos }^2\theta +\cos \theta -3=0$

This is a quadratic equation in $\cos \theta .$
$\Rightarrow 10{\cos }^2\theta +(6-5)\cos \theta -3=0$
$\Rightarrow 10{\cos }^2\theta +6\cos \theta -5\cos \theta -3=0$
$\Rightarrow 2\cos \theta (5\cos \theta +3)-(5\cos \theta +3)=0$
$\Rightarrow (2\cos \theta -1)(5\cos \theta +3)=0$

$\Rightarrow 2\cos \theta -1=0\text{or}5\cos \theta +3=0.$

For the first case:
$\cos \theta =\frac{1}{2}=\cos \frac{\pi }{3}.$
$\Rightarrow \theta =2n\pi \pm \frac{\pi }{3},n\in Z\cdots \left(A\right)$

For the second case:
$\cos \theta =-\frac{3}{5}<0$
$\therefore$ There exist an angle $\alpha$ such that $\cos \alpha =-\frac{3}{5}.$

$\Rightarrow \cos \theta =\cos \alpha$ is $\theta =2n\pi \pm \alpha ,n\in Z\cdots \left(B\right)$

The complete general solution is $A\cup B.$
$\Rightarrow \theta \in \{2n\pi \pm \frac{\pi }{3}\}\bigcup \{2n\pi \pm \alpha \},n\in Z,$ where $\cos \alpha =-\frac{3}{5}.$