The least value of
$\cos^2\theta - 6\sin\theta\cos\theta + 3\sin^2\theta + 2$
is:
🎥 Video solution / Text Solution of this question is given below:
$f(\theta) = \frac{1+\cos2\theta}{2} - 3\sin2\theta + 3\left(\frac{1-\cos2\theta}{2}\right) + 2$
$= 4 - 3\sin2\theta - \cos2\theta$
Minimum of $a\sin x + b\cos x = \sqrt{a^2 + b^2}$
$\Rightarrow \min f = 4 - \sqrt{(-3)^2 + (-1)^2} = 4 - \sqrt{10}$