#### Answer

The solution of this system is unique and it is:
$$x=\cos\theta-\sin\theta,\quad y= \cos\theta + \sin\theta.$$

#### Work Step by Step

Here we can treat sines and cosines as ordinary coefficients and we will solve this system as we would solve any system of two equations with two unknowns.
Step 1: Multiply the first equation by $\sin\theta$ and the second equation by $\cos\theta$:
\begin{align*}
(\sin\theta\cos\theta)x+(\sin^2\theta)y=&\sin\theta\\
(-\sin\theta\cos\theta)x+(\cos^2\theta)y =& \cos\theta.
\end{align*}
Step 2: Add these two equations together to eliminate $x$ and find $y$:
$$(\sin\theta\cos\theta)x +(-\sin\theta\cos\theta)x + (\sin^2\theta)y +(\cos^2\theta)y = \sin\theta+\cos\theta$$
which becomes
$$y(\sin^2\theta+\cos^2\theta) = \sin\theta+\cos\theta.$$
Remember that for any $\theta$ we have that $\sin^2\theta+\cos^2\theta = 1$ so this simplifies to
$$y=\sin\theta + \cos\theta$$
Step 3: Now let us do the similar thing as in step 1. Multiply the first equation by $\cos\theta$ and the second one by $\sin\theta$:
\begin{align*}
(\cos^2\theta)x+ (\sin\theta\cos\theta) y = &\cos\theta\\
(-\sin^2\theta)x+(\sin\theta\cos\theta)y = &\sin\theta
\end{align*}
Step 4: Now subtract these equations to eliminate $y$ and find $x$ :
$$(\cos^2\theta)x-(-\sin^2\theta)x + (\sin\theta\cos\theta) y -(\sin\theta\cos\theta)y =\cos\theta- \sin\theta$$
This then gives
$$x(\sin^2\theta+\cos^2\theta) = \cos\theta-\sin\theta.$$
Using the fact that $\sin^2\theta +\cos^2\theta = 1$ we again get simpler expression:
$$x=\cos\theta-\sin\theta.$$