Step 1: Use Multiple Angles:
Let $\theta=36^{\circ}$ so $5\theta=180^{\circ}$ and $3\theta=180^{\circ}-2\theta$. Hence $\cos3\theta=-\cos2\theta$.
Step 2: Expand:
With $c=\cos\theta$: $4c^3-3c=-(2c^2-1)$, so $4c^3+2c^2-3c-1=0$. The value $c=-1$ satisfies it, so factor as $(c+1)(4c^2-2c-1)=0$.
Step 3: Solve:
From $4c^2-2c-1=0$: $c=\dfrac{1\pm\sqrt5}{4}$. The cosine of $36^{\circ}$ is positive, so $c=\dfrac{1+\sqrt5}{4}$. Option (C).
Final Answer:
Option (C).
\[ \boxed{\text{(C) } \frac{1+\sqrt{5}}{4}} \]