1. The problem is to check if $3\sin 2\theta = 3\sin^2 \theta$. 2. Recall the double-angle identity: $\sin 2\theta = 2\sin \theta \cos \theta$. 3. Substitute into the left side: $3\sin 2\theta = 3 \times 2\sin \theta \cos \theta = 6\sin \theta \cos \theta$. 4. The right side is $3\sin^2 \theta$. 5. So, $3\sin 2\theta = 6\sin \theta \cos \theta$ and $3\sin^2 \theta = 3(\sin \theta)^2$. 6. These two expressions are generally not equal because $6\sin \theta \cos \theta \neq 3\sin^2 \theta$. 7. Conclusion: $3\sin 2\theta \neq 3\sin^2 \theta$ in general.