To solve the equation \( \cos 6\theta = \sin 3\theta \), we use the identity that relates sine and cosine: \( \cos \alpha = \sin (90^\circ - \alpha) \). Applying this, we get: \[\cos 6\theta = \sin 3\theta \Rightarrow \sin 3\theta = \sin (90^\circ - 6\theta)\] Since \(\sin A = \sin B\), this implies that: \[3\theta = 90^\circ - 6\theta + n \times 180^\circ\] for \(n \in \mathbb{Z}\). Simplifying, we have: \[3\theta + 6\theta = 90^\circ + n \times 180^\circ\] \[9\theta = 90^\circ + n \times 180^\circ\] Dividing through by 9 gives: \[\theta = 10^\circ + n \times 20^\circ\] Now, to find specific solutions for \(\theta\) within the range of conventional angles, choose \(n = 0\), giving \(\theta = 10^\circ\). This is a valid solution that aligns with the provided options.