$\def\t#1{\theta_{#1}}$
\begin{align} \sin\t1+\sin\t2&=c_1 \tag{1}\label{1} \\
\cos\t1+\cos\t2&=c_2 \tag{2}\label{2}
\\ \sin(\t2-\t1) &= c_1\cos\t1
- c_2\sin\t1 \tag{3}\label{3}
\end{align}
The sum of squares of \eqref{2} and \eqref{1}
gives
\begin{align}
\cos^2\t1&+2\,\cos\t1\cos\t2+\cos^2\t2
+
\sin^2\t1+2\,\sin\t1\sin\t2+\sin^2\t2
=
c_1^2+c_2^2,
\\
\cos\t1\cos\t2&+\sin\t1\sin\t2
=
\cos(\t2-\t1)
=\tfrac12(c_1^2+c_2^2)-1
.
\end{align}
hence
\begin{align}
\sin(\t2-\t1)&=\sqrt{1-
\left(
\tfrac12(c_1^2+c_2^2)-1
\right)^2}
\end{align}
Next, $\eqref{1}\times \sin\t1+\eqref{2}\times \cos\t1$
results in
\begin{align}
1+\cos(\t2-\t1)
&=
c_1\,\sin\t1+c_2\,\cos\t1
\tag{4}\label{4}
\end{align}
Equations \eqref{4} and \eqref{3}
present a $2\times2$ linear system
\begin{align}
\begin{bmatrix}
\phantom{-}c_1&c_2\\ -c_2&c_1
\end{bmatrix}
\begin{bmatrix}
\sin\t1\\ \cos\t1
\end{bmatrix}
&=
\begin{bmatrix}
\tfrac12(c_1^2+c_2^2)
\\
\sqrt{1-
\left(
\tfrac12(c_1^2+c_2^2)-1
\right)^2}
\end{bmatrix}
.
\tag{5}\label{5}
\end{align}
Unless $c_1,c_2$ are both zero (hence $\t2 = \t1-\pi$),
solution to \eqref{5}
provides us with $\sin\t1$ and $\cos\t1$
(check that $\sin^2\t1 + \cos^2\t1\equiv1$).
Together with \eqref{1} and \eqref{2}
we can get the $\sin$ and $\cos$ of the other angle.
Also, hence the system is symmetric,
the solution is either $\{\t1,\t2\}$ or $\{\t2,\t1\}$.
Best Answer
Rather, multiply the first equation by $\cos q$ and the second by $-\sin q$ and add these scaled equations to obtain $$x\cos(p+q)-y\sin(p+q)+z(\cos q -\sin q)=\cos q -\sin q+1.$$ Since from the last of the original triple we know that $x\cos(p+q)-y\sin(p+q)=2-z,$ we can substitute in the above sum and letting $K=\cos q -\sin q,$ we obtain $$2-z+Kz=K+1,$$ which you may now substitute in any pair of the original system to eliminate $z,$ so that you now have a pair in $x$ and $y$ alone. I believe you can find the way from there.