Aspire Faculty ID #18023 · Topic: JEE Main 2026 (23 January Evening Shift) · Just now
JEE Main 2026 (23 January Evening Shift)

The least value of

$\cos^2\theta - 6\sin\theta\cos\theta + 3\sin^2\theta + 2$

is:

Solution

$f(\theta) = \frac{1+\cos2\theta}{2} - 3\sin2\theta + 3\left(\frac{1-\cos2\theta}{2}\right) + 2$

$= 4 - 3\sin2\theta - \cos2\theta$

Minimum of $a\sin x + b\cos x = \sqrt{a^2 + b^2}$

$\Rightarrow \min f = 4 - \sqrt{(-3)^2 + (-1)^2} = 4 - \sqrt{10}$

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