Question:

The least value of $(\cos^2\theta-6\sin\theta\cos\theta+3\sin^2\theta+2)$ is

-1

4-√10

4+√10

Step-by-Step Solution

Key Concept: Rewrite: $\tfrac{1+\cos2\theta}{2}+\tfrac{3(1-\cos2\theta)}{2}-3\sin2\theta+2=4-\cos2\theta-3\sin2\theta$.

Least value $=4-\sqrt{10}$.

Correct Answer: 3