 Solution

A cyclic group of order $p$ contains an element of order two if and only if $2$ divides $p$. Necessity follows from Lagrange's theorem; for even $p$, the unique such element is $c=g_1^{p/2}$. HenceBecause $g_1$ has order $p>2$ and the representation is faithful, $D(g_1)$ cannot be a reflection, which has order two. It is therefore a rotation through a primitive angle $\vartheta =2\pi q/p$, with $\gcd (q,p)=1$. The condition on the greatest common divisor is exactly what makes its order $p$ rather than a proper divisor. Since $p$ is even, $q$ is odd, so $D(c)$ is the half-turn:The character is the trace:Choosing the primitive generator with $q=1$ gives the customary rotation representation. The general coprime value is included because faithfulness alone does not select that generator. These are the two-dimensional cases in the irreducible real representations of a finite cyclic group.