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digimon story time stranger release date - Richter Guitar

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 But use a known result: A sum of two sinusoids with incommensurate frequencies has approximately $ 2n $ zeros per cycle, but here we seek level crossings.

📌 Instead, consider that $ f(\theta) = \sin(3\theta) + \cos(4\theta) $ is differentiable, and $ f'(\theta) = 3\cos(3\theta) - 4\sin(4\theta) $. The number of solutions to $ |f(\theta)| = 1 $ is discrete and finite in any bounded interval if we consider level sets — but actually, it's continuous, so it crosses $ \pm1 $ infinitely often? Wait: $ \theta \in [0,7\pi) $ is infinite, but the problem likely assumes one full cycle of the group behavior — but no time bound was given.

📌 Wait: Re-examining the original question — it says "in one full cycle of $ S $", but $ S(t) $ has period $ \text{LCM}(14/3, 7/2) $. Compute:

📌 $ \frac{14}{3} = \frac{28}{6} $, $ \frac{7}{2} = \frac{21}{6} $, LCM of periods $ \frac{14}{3} $ and $ \frac{7}{2} $: write as $ \frac{28}{6}, \frac{21}{6} $, GCD of 28 and 21 is 7, so period is $ \frac{14}{\gcd(14,21)} \cdot \frac{1}{\text{some factor}} $. Better: find least $ T $ such that $ \frac{3\pi}{7}T = 2\pi m $, $ \frac{4\pi}{7}T = 2\pi n $, so $ T = \frac{14m}{3} = \frac{7n}{2} \Rightarrow \frac{14m}{3} = \frac{7n}{2} \Rightarrow 28m = 21n \Rightarrow 4m = 3n $. Small

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