Should you find theta?

First we can find a recurrence relation for $x_n$ . Indeed, since $x_n=\frac {(2e^{i\theta})^n+(2e^{-i\theta})^n}{2},$ then $x_n$ satisfies a linear recurrence relation with characteristic polynomial $(r-2e^{i\theta})(r-2e^{-i\theta})=$ $=r^2- 4r\cos \theta +4$ . For details you can see the last part of the wiki Linear Recurrence Relations with repeated roots . Then this implies that the sequence satisfies the recurrence: $x_n=4(\cos \theta)x_{n-1}-4x_{n-2}\quad \text{for all}\quad n\geq 2.$

Making $n=3$ in the recurrence, substituting $x_3$ and $x_2$ by the given values, and $x_1$ by $2\cos\theta,$ we obtain the equation $-\frac{184}{27}=4(\cos\theta) (-\frac{28}{9})-8\cos\theta.$ Solving the equation we obtain that $\cos\theta=\frac{1}{3}.$ Now, making $n=4$ in the recurrence and substituting the $\cos \theta$ we obtain that $x_4=4(\cos \theta)x_3-4x_2=4(\frac{1}{3})(-\frac{184}{27})-4(-\frac{28}{9})=\frac{272}{81}.$
Then the solution to the problem is $\boxed{353}.$

$x_{2} =2^2 cos(2θ)=-28/9 \Rightarrow cos(2θ)=-7/9$ Now, $x_{4} =2^4(cos(4θ)),=16*(2cos^2(2θ)-1)=16(2(-7/9)^2 -1)=16*(17/81)$ 272+81=353

$x_{3}$ is actually not required.

From question, $\large x_2=-\frac{28}{9}$ = $\large 2^2 \cos(2\theta)$ .

We note that $\large x_4=2^4 \cos(4\theta)$ = $\large 2^4 (2(cos(2\theta))^2$ -1)). Substitute and simplify.