1. State the problem: Solve for x given $\cos(3x+6^\circ)=\tfrac{1}{2}$ and $3x+6^\circ$ is an acute angle. 2. Understand the acute-angle constraint: an acute angle means it lies strictly between $0^\circ$ and $90^\circ$. 3. General solutions for cosine: $\cos\theta=\tfrac{1}{2}$ when $\theta=60^\circ+360^\circ k$ or $\theta=300^\circ+360^\circ k$ for integer $k$. 4. Apply the acute constraint: among the general solutions, only $60^\circ$ lies in $0^\circ<\theta<90^\circ$; all other values are outside this range. 5. Set $3x+6^\circ=60^\circ$ and solve: $3x=54^\circ$. 6. Divide by 3 to find $x=18^\circ$. 7. Check: $3(18^\circ)+6^\circ=60^\circ$ and $\cos60^\circ=\tfrac{1}{2}$. Final answer: $x=18^\circ$.