\min |2\sin x + 3\cos x + 4| = \min (2\sin x + 3\cos x + 4) = 4 - \sqrt{13} + 0? \quad \text{Wait — no: the minimum of the inner expression is $ 4 - \sqrt{13} \approx 0.395 $, so the minimum of the absolute value is just its minimum value because it's positive.

February 22, 2026

$\min |2\sin x + 3\cos x + 4| = \min (2\sin x + 3\cos x + 4) = 4 - \sqrt{13} + 0? \quad \text{Wait — no: the minimum of the inner expression is $ 4 - \sqrt{13} \approx 0.395 $, so the minimum of the absolute value is just its minimum value because it's positive.$

Since $ \sqrt{13} \approx 3.605 $, $ 4 - \sqrt{13} \approx 0.395 > 0 $, so the expression never reaches zero.
Therefore, the absolute value $ |2\sin x + 3\cos x + 4| $ achieves its minimum at the minimum of the continuous function on an interval where it is always positive.
The minimum occurs when $ 2\sin x + 3\cos x $ is minimized, but since it's always greater than $ -\sqrt{13} $, and greater than $ -4 $, the expression inside is always positive.
\min |2\sin x + 3\cos x + 4| = 4 - \sqrt{13} \quad \text{if } 4 - \sqrt{13} > 0
But $ 4 - \sqrt{13} > 0 $ since $ \sqrt{13} < 4 $, so yes.
But wait: is this the minimum? Since the expression inside is always greater than $ 0 $, the absolute value is minimized exactly when $ 2\sin x + 3\cos x $ is minimized.

\min |2\sin x + 3\cos x + 4| = \min (2\sin x + 3\cos x + 4) = 4 - \sqrt{13} + 0? \quad \text{Wait — no: the minimum of the inner expression is $ 4 - \sqrt{13} \approx 0.395 $, so the minimum of the absolute value is just its minimum value because it's positive.

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