Question

What is the general solution of $2{\cos }^2\theta +1=3\cos \theta ?$

Answer 1

Given that $2{\cos }^2\theta +1=3\cos \theta$
$\Rightarrow 2{\cos }^2\theta -3\cos \theta +1=0$
$\Rightarrow 2{\cos }^2\theta -2\cos \theta -\cos \theta +1=0$
$\Rightarrow 2\cos \theta (\cos \theta -1)-1(\cos \theta -1)=0$
$\Rightarrow (2\cos \theta -1)(\cos \theta -1)=0$
$\Rightarrow (2\cos \theta -1)=0\text{or}(\cos \theta -1)=0$
$\Rightarrow \cos \theta =\frac{1}{2}\text{or}\cos \theta =1$
$\therefore \cos \theta =\cos \frac{\pi }{3}\text{or}\cos \theta =\cos 0$
Using $\cos \theta =\cos \alpha \Rightarrow \theta =2n\pi \pm \alpha ,n\in Z$
$\Rightarrow \theta =2n\pi \pm \frac{\pi }{3};n\in Z\text{or}\theta =2m\pi \pm 0;m\in Z$
i.e. $\theta \in \{2n\pi \pm \frac{\pi }{3};n\in Z\}\cup \{2m\pi ;m\in Z\}$