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\( a + b + c + d = 4 \) - Ready Digital AB

📅 March 7, 2026 👤 scraface

Mar 07, 2026

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where $ R = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} $.
2\sin x + 3\cos x = \sqrt{13} \sin(x + \phi)
The range of $ \sqrt{13} \sin(x + \phi) $ is $ [-\sqrt{13}, \sqrt{13}]...
2\sin x + 3\cos x + 4 \in [4 - \sqrt{13}, 4 + \sqrt{13}]
Since $ \sqrt{13} \approx 3.605 $, $ 4 - \sqrt{13} \approx 0.395 > 0 $...
Therefore, the absolute value $ |2\sin x + 3\cos x + 4| $ achieves its...
The minimum occurs when $ 2\sin x + 3\cos x $ is minimized, but since...
\min |2\sin x + 3\cos x + 4| = \min (2\sin x + 3\cos x + 4) = 4 - \sqr...
\min |2\sin x + 3\cos x + 4| = 4 - \sqrt{13} \quad \text{if } 4 - \sqr...
But $ 4 - \sqrt{13} > 0 $ since $ \sqrt{13} < 4 $, so yes.

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