If $\theta \in [0,2\pi ]$ and $\cos 2\theta =\cos \theta +\sin \theta$ then the sum of all values of $\theta$ satisfying the equation is

Given, $\cos 2\theta =\cos \theta +\sin \theta$

Using, $\cos 2A={\cos }^{2}x-{\sin }^{2}x$, we get

${\cos }^2\theta -{\sin }^2\theta =\cos \theta +\sin \theta$

$\Rightarrow \left(\cos \theta +\sin \theta \right)\left(\cos \theta -\sin \theta \right)=\cos \theta +\sin \theta$  $\left[\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right]$

$\Rightarrow \left(\cos \theta +\sin \theta \right)\left(\cos \theta -\sin \theta -1\right)=0$

$\Rightarrow \cos \theta +\sin \theta =0$ or $\cos \theta -\sin \theta =1$

Now, $\cos \theta +\sin \theta =0\Rightarrow \cos \theta =-\sin \theta$

$\Rightarrow \tan \theta =-1$

$\Rightarrow \theta =\frac{3\pi }{4}, \frac{7\pi }{4}$

And, $\cos \theta -\sin \theta =1$

Either $\cos \theta =1, \sin \theta =0$

$\Rightarrow \theta =0, 2\pi$

Or $\cos \theta =0$, $\sin \theta =-1$

$\Rightarrow \theta =\frac{3\pi }{2}$

Now, Sum of all possible values of $\theta$ is $\frac{3\pi }{4}+\frac{7\pi }{4}+0+2\pi +\frac{3\pi }{2}=\frac{24\pi }{4}$