Inspired by Mridul Jain

Using the identities $\sin x +\sin y=2\sin (\frac{x+y}{2})\cos(\frac{x-y}{2})$ and $\cos x +\cos y=2\cos (\frac{x+y}{2})\cos(\frac{x-y}{2}),$ we can obtain $\sin {n\theta} +\sin {(n-2)\theta}=2\sin \frac{(n-1)\theta}{2}\cos{\theta}$ and $\cos {n\theta} +\cos {(n-2)\theta}=2\cos \frac{(n-1)\theta}{2}\cos{\theta},$ respectively. From these two equations it is easy to get the recurrence : $x_n=2(\cos \theta) x_{n-1}-x_{n-2}.$ By making $n=3$ into the recurrence and using the initial values given in the problem, we obtain the equation $-12=2(\cos\theta)(3)-15.$ So $\cos\theta=\frac{1}{2},$ and, therefore, the recurrence becomes $x_n=x_{n-1}-x_{n-2}.$

Now, the rest is very easy: $x_4=x_3-x_2=-12-3=-15$ $x_5=x_4-x_3=-15-(-12)=-3$ $x_6=x_5-x_4=-3-(-15)=12$ $x_7=x_6-x_5=12-(-3)=15$ So the answer is $\boxed{15}.$

Adding $x_1$ and $x_3$ we get, $2asin(2\theta)\times cos(\theta) + 2bcos(2\theta)\times cos(\theta)$ =3.

$\implies$ $2cos(\theta) [x_2]$ =3.

$\implies$ $2cos(\theta)=1$

$\theta$ = $2n\pi \pm \frac {\pi}{3}$ {n belongs to integers}

$7\theta$ = $14n\pi \pm \frac{7\pi}{3}$

$7\theta$ = $2\pi (7n \pm 1) \pm \frac{\pi}{3}$

Since sin and cos are periodic about $2\pi$ , $7\theta=\theta$

$therefore$ $x_7 =x_1 =15$ .