Step 1: Spot the structure:
For any $2 \times 2$ matrix, the sum $A + A^{\prime}$ has twice each diagonal entry on the diagonal.
The off-diagonal entries become $a_{12} + a_{21}$.
Here $a_{12} = -\sin x$ and $a_{21} = \sin x$, so they add to $0$.
Step 2: Reduce the condition:
So $A + A^{\prime} = 2\cos x\,I$.
Setting this equal to $I$ means the scalar must be $1$: $2\cos x = 1$.
Step 3: Test each option in $2\cos x$:
For (A), $x = 0$: $2\cos 0 = 2$. This is not $1$.
For (B), $x = \dfrac{\pi}{4}$: $2\cdot\dfrac{\sqrt{2}}{2} = \sqrt{2}$. This is not $1$.
For (C), $x = \dfrac{\pi}{3}$: $2\cdot\dfrac{1}{2} = 1$. This works.
For (D), $x = \dfrac{\pi}{2}$: $2\cdot 0 = 0$. This is not $1$.
Step 4: Conclude:
Only $x = \dfrac{\pi}{3}$ makes the scalar equal to $1$.
Final Answer:
The correct choice is option (C). \[ \boxed{x = \dfrac{\pi}{3}} \]