Solve the following expression, $\cos \theta +\cos 3\theta -2\cos 2\theta =0$.
A. $\theta =\left( 2n+1 \right)\dfrac{\pi }{4},n\in z$
B. $\theta =m\pi ,m\in z$
C. $\theta =2m\pi ,m\in z$
D. $\theta =\left( 2n+1 \right)\dfrac{\pi }{8},n\in z$

Hint: To solve this question, we should know a few trigonometric identities like, $\cos a+\cos b=2\cos \dfrac{a+b}{2}\cos \dfrac{a-b}{2}$. We should also have some knowledge of general equations like if $\cos x=\cos \alpha $, then $x=2m\pi \pm \alpha $, where$\alpha \in \left[ 0,\dfrac{\pi }{2} \right]$ and if $\cos x=0$, then $x=n\pi +\dfrac{\pi }{2}$. Also, we should know that $\cos 0=1$. We can solve this question by using these properties.

Complete step-by-step answer:
In this question, we have been asked to solve an equation, that is, $\cos \theta +\cos 3\theta -2\cos 2\theta =0$. To solve this question, we will write the given equation as follows,
$\cos \theta +\cos 3\theta -2\cos 2\theta =0$
Now, we know that $\cos a+\cos b=2\cos \dfrac{a+b}{2}\cos \dfrac{a-b}{2}$. So, for $a=3\theta $ and $b=\theta $, we will the equation as,
$2\cos \left( \dfrac{3\theta +\theta }{2} \right)\cos \left( \dfrac{3\theta -\theta }{2} \right)-2\cos 2\theta =0$
And we can write it further as,
$\begin{align}
  & 2\cos \left( \dfrac{4\theta }{2} \right)\cos \left( \dfrac{2\theta }{2} \right)-2\cos 2\theta =0 \\
 & \Rightarrow 2\cos 2\theta \cos \theta -2\cos 2\theta =0 \\
\end{align}$
And, we can see that $2\cos 2\theta $ can be taken out as the common term. So, we get the above equation as,
$2\cos 2\theta \left( \cos \theta -1 \right)=0$
Now, we know that the equation has to satisfy either $2\cos 2\theta =0$ or $\left( \cos \theta -1 \right)=0$. So, we can write as, $\cos 2\theta =0$ or $\cos \theta =1$. We know that $\cos 0=1$, so we can write the above equation as, $\cos 2\theta =0$ or $\cos \theta =\cos 0$.
Now, we know that if $\cos \theta =\cos \alpha $, then $\theta =2m\pi \pm \alpha $ and if $\cos \theta =0$, then $\theta =n\pi +\dfrac{\pi }{2}$, so,
$2\theta =n\pi \pm \dfrac{\pi }{2}$ or $\theta =2m\pi \pm 0$
And, we can further write it as,
$\begin{align}
  & \theta =\dfrac{n\pi }{2}\pm \dfrac{\pi }{4}\text{ }or\text{ }\theta =2m\pi ;m,n\in z \\
 & \Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{4}\text{ }and\text{ }\theta =2m\pi ;m,n\in z \\
\end{align}$
Hence, we can say that according to the options given in the question, options A and C are the correct options for this question.

Note: We have to remember while solving this question that, if $\cos \theta =0$, then $\theta =n\pi +\dfrac{\pi }{2}$. It is because if we write $0=\cos \dfrac{\pi }{2}$ and then write $\cos \theta =\cos \dfrac{\pi }{2}$ and then use the general formula of, if $\cos \theta =\cos \alpha $, then $\theta =2m\pi \pm \alpha $, we may skip a few values and will get the incorrect answers.