Question: A mammalogist observes that the social cohesion index $ S $ of a primate group over time is modeled by $ S = \left| \sin\left(\frac{3\pi}{7} t\right) + \cos\left(\frac{4\pi}{7} t\right) \right| $, where $ t $ is time in days. How many times in one full cycle of $ S $ (i.e., over $ t \in [0, 7) $) does $ S = 1 $?

["Understanding When Social Cohesion Reaches Maximum: A Deep Dive into $ S = \left| \sin\left(\frac{3\pi}{7} t\right) + \cos\left(\frac{4\pi}{7} t\right) \right| $", "A primate group’s social cohesion is quantified by the function:\n$$\nS(t) = \left| \sin\left(\frac{3\pi}{7} t\right) + \cos\left(\frac{4\pi}{7} t\right) \right|\n$$\nwhere $ t $ represents time in days. The observable peak of cohesion occurs when $ S(t) = 1 $, a condition when the combined influence of the sine and cosine terms reaches its maximum absolute amplitude.", "To determine how many times this maximum occurs in one full cycle, we analyze the periodicity and behavior of the function over one complete period $ t \in [0, 7) $.", "---", "### Step 1: Identify the Period of $ S(t) $", "The function involves two trigonometric components:\n- $ \sin\left(\frac{3\pi}{7} t\right) $ has period $ \frac{2\pi}{\frac{3\pi}{7}} = \frac{14}{3} $\n- $ \cos\left(\frac{4\pi}{7} t\right) $ has period $ \frac{2\pi}{\frac{4\pi}{7}} = \frac{7}{2} $", "The overall period of $ S(t) $ is the least common multiple (LCM) of $ \frac{14}{3} $ and $ \frac{7}{2} $. To compute this:", "Convert to fractions:\n- $ \frac{14}{3} = \frac{28}{6} $\n- $ \frac{7}{2} = \frac{21}{6} $", "We find the LCM of periods using harmonic analysis. Alternatively, observe that both functions complete full cycles over multiples of $ 14 $ (since LCM of $ \ ext{period}_1 $ and $ \ ext{period}_2 $ in denominator leads to common period $ 14 $):", "Check: $ \sin\left(\frac{3\pi}{7}(t + 14)\right) = \sin\left(\frac{3\pi}{7}t + 6\pi\right) = \sin\left(\frac{3\pi}{7}t\right) $\n$ \cos\left(\frac{4\pi}{7}(t + 14)\right) = \cos\left(\frac{4\pi}{7}t + 8\pi\right) = \cos\left(\frac{4\pi}{7}t\right) $", "Thus, $ S(t) $ is periodic with period $ 14 $. However, the question specifies one full cycle as $ t \in [0, 7) $. So we analyze behavior over half a full cycle — but note that $ S(t) $ is symmetric and may achieve peaks within this interval.", "But wait: although the full period is 14, the function $ S(t) $ may complete multiple full cycles of fluctuation within $ [0, 7) $. Let’s instead reframe: since the individual frequencies are rational multiples of $ \pi $, the function $ S(t) $ is almost periodic, and over $ [0, 7) $, we can determine how many times $ |f(t)| = 1 $.", "But more directly: $ S(t) = 1 $ when\n$$\n\left| \sin\left(\frac{3\pi}{7} t\right) + \cos\left(\frac{4\pi}{7} t\right) \right| = 1\n\Rightarrow \sin\left(\frac{3\pi}{7} t\right) + \cos\left(\frac{4\pi}{7} t\right) = \pm 1\n$$", "Let $ \omega = \frac{\pi}{7} $, so the arguments become:\n- $ \frac{3\omega}{1} t = 3\omega t $\n- $ \frac{4\omega}{1} t = 4\omega t $", "Thus:\n$$\nS(t) = \left| \sin(3\omega t) + \cos(4\omega t) \right|, \quad \omega = \frac{\pi}{7}\n$$", "Now, define $ f(t) = \sin(3\omega t) + \cos(4\omega t) $. We seek the number of solutions to $ |f(t)| = 1 $ for $ t \in [0, 7) $, noting $ 7\omega = \pi $, so $ t = 7 $ corresponds to $ 7\omega \cdot t = 7\pi \Rightarrow $ frequency envelope tied to $ \pi $.", "But observe: since $ \omega = \pi/7 $, then:\n- $ 3\pi/7 \cdot t = 3\omega t $\n- $ 4\pi/7 \cdot t = 4\omega t $", "The functions have angular frequencies 3 and 4, both integer multiples of $ \omega $. Thus $ f(t) $ is periodic with period $ \ ext{LCM}(2\pi/3\omega, 2\pi/4\omega) = \ ext{LCM}(2\pi/(3\omega), 2\pi/(4\omega)) = \frac{2\pi}{\omega} \cdot \ ext{LCM}(1/3, 1/4) $", "Better: the overall period is $ \frac{2\pi}{\gcd(3,4)\omega} = \frac{2\pi}{\omega} = 7 $, since 3 and 4 are coprime.", "So $ f(t) $ is periodic with period 7. Thus $ S(t) = |f(t)| $ is periodic with period 7.", "Therefore, we analyze $ S(t) $ over $ [0, 7) $ to capture one full cycle.", "---", "### Step 2: Analyze When $ |f(t)| = 1 $", "We want the number of times $ |f(t)| = 1 $ for $ t \in [0, 7) $, where $ f(t) = \sin(3\omega t) + \cos(4\omega t) $, $ \omega = \pi/7 $", "Let $ x = \omega t = \frac{\pi}{7} t $. As $ t $ goes from 0 to 7, $ x \in [0, \pi] $", "So define $ g(x) = \sin(3x) + \cos(4x) $, $ x \in [0, \pi] $, and solve $ |g(x)| = 1 $", "This is a smooth function. We seek the number of solutions to $ g(x) = 1 $ or $ g(x) = -1 $ in $ [0, \pi] $", "---", "### Step 3: Use Trigonometric Identity to Simplify", "Try to write $ g(x) = \sin(3x) + \cos(4x) $ in a more manageable form.", "Use identities:\n- $ \sin(3x) = 3\sin x - 4\sin^3 x $\n- $ \cos(4x) = 1 - 2\sin^2(2x) $, but this may not help.", "Alternatively, consider Fourier components. The function $ g(x) $ is almost a sum of harmonics. But instead, analyze graphically and analytically.", "Note: $ g(x) = \sin(3x) + \cos(4x) $ is a real trigonometric polynomial. The equation $ |g(x)| = 1 $ defines level sets.", "We analyze the number of solutions to $ g(x) = 1 $ and $ g(x) = -1 $ in $ [0, \pi] $. Each equation can have multiple solutions.", "Let’s estimate the behavior:", "- $ \sin(3x) $: period $ 2\pi/3 \approx 2.09 $, oscillates 3 times in $ [0, 2\pi) $, so in $ [0, \pi] $ it completes $ 3/2 = 1.5 $ cycles → 1 full hump up, down, and a partial one.", "- $ \cos(4x) $: period $ 2\pi/4 = \pi/2 $, so 2 full cycles in $ [0, \pi] $", "Thus, $ g(x) $ is a combination of fast (4) and slow (3) oscillations. Their sum will be complex, but maximum possible amplitude?", "Max of $ |\sin(3x) + \cos(4x)| \leq 1 + 1 = 2 $, so $ |g(x)| = 1 $ is attainable.", "Let’s find critical points by differentiating:\n$$\ng'(x) = 3\cos(3x) - 4\sin(4x)\n$$\nSet $ g'(x) = 0 $:\n$$\n3\cos(3x) = 4\sin(4x)\n$$\nUse $ \sin(4x) = 2\sin(2x)\cos(2x) = 4\sin x \cos x (1 - 2\sin^2 x) $, messy.", "Instead, consider numerical sampling and symmetry.", "---", "### Step 4: Sample Key Points", "Evaluate $ g(x) = \sin(3x) + \cos(4x) $ at key points in $ [0, \pi] $", "| $ x $ (rad) | $ 3x $ | $ \sin(3x) $ | $ 4x $ | $ \cos(4x) $ | $ g(x) $ |\n|-------------|--------|---------------|--------|----------------|-----------|\n| 0 | 0 | 0 | 0 | 1 | 1 |\n| $ \pi/6 $ | $ \pi/2 $ | 1 | $ 2\pi/3 $ | $ \cos(2\pi/3) = -0.5 $ | 0.5 |\n| $ \pi/4 $ | $ 3\pi/4 $ | $ \sin(135^\circ) = \frac{\sqrt{2}}{2} \approx 0.707 $ | $ \pi $ | $ \cos(\pi) = -1 $ | $ -0.293 $ |\n| $ \pi/3 $ | $ \pi $ | 0 | $ 4\pi/3 $ | $ \cos(240^\circ) = -0.5 $ | -0.5 |\n| $ \pi/2 $ | $ 3\pi/2 $ | -1 | $ 2\pi $ | 1 | 0 |\n| $ 2\pi/3 $ | $ 2\pi $ | 0 | $ 8\pi/3 \equiv 2\pi/3 $ | $ \cos(240^\circ) = -0.5 $ | -0.5 |\n| $ 3\pi/4 $ | $ 9\pi/4 \equiv \pi/4 $ | $ \sin(\pi/4) \approx 0.707 $ | $ 3\pi $ | $ \cos(3\pi) = -1 $ | -0.293 |\n| $ 5\pi/6 $ | $ 5\pi/2 \equiv \pi/2 $ | 1 | $ 10\pi/3 \equiv 4\pi/3 $ | $ -0.5 $ | 0.5 |\n| $ \pi $ | $ 3\pi $ | 0 | $ 4\pi $ | 1 | 1 |", "We see $ g(x) = 1 $ at $ x = 0 $, $ x = \pi $", "Now check where $ g(x) = -1 $:\nAt $ x = \pi/6 $: $ g \approx 0.5 $\nAt $ x = 5\pi/6 $: $ g = 0.5 $ — both positive.", "Try $ x = 2\pi/3 $: $ g = -0.5 $, $ x = \pi/3 $: $ -0.5 $ — no negative 1.", "But wait: $ \sin(3x) \geq -1 $, $ \cos(4x) \geq -1 $, so $ g(x) \geq -2 $. Can it reach -1?", "Try $ x = \frac{2\pi}{3} + \epsilon $. Try numerical solver.", "Alternatively, note that $ g(x) $ is continuous and piecewise smooth. From the values:\n- $ g(0) = 1 $\n- Increases to $ x \approx 0.5 $? Try $ x = 0.4 $:\n $ 3x = 1.2 $ rad ≈ 68.7° → $ \sin \approx 0.93 $\n $ 4x = 1.6 $ rad ≈ 91.7° → $ \cos \approx -0.02 $ → $ g \approx 0.91 $", "At $ x = 0.3 $: $ 3x = 0.9 $ rad ≈ 51.6° → $ \sin \approx 0.79 $\n$ 4x = 1.2 $ rad ≈ 68.7° → $ \cos \approx 0.37 $ → $ g \approx 1.16 > 1 $", "So $ g(x) > 1 $ near 0.\nAt $ x = 0.5 $: $ 3x = 1.5 $ rad ≈ 85.9° → $ \sin \approx 0.997 $\n$ 4x = 2.0 $ rad ≈ 114.6° → $ \cos \approx -0.41 $ → $ g \approx 0.587 $", "So $ g(x) $ starts at 1, rises slightly, then falls to 0.5 at $ \pi/6 $, dips to -0.3 at $ \pi/4 $, reaches $-0.5$ at $ \pi/3 $, stays negative, then rises back to 0 at $ 5\pi/6 $, then to 1 at $ \pi $.", "So $ g(x) = 1 $ at $ x = 0 $ and $ x = \pi $ — and possibly again? But at $ x = \pi $, $ 3x = 3\pi $, $ \sin = 0 $, $ 4x = 4\pi $, $ \cos = 1 $ → indeed $ g(\pi) = 1 $", "Now, is there a point where $ g(x) = -1 $? Try $ x = 2.5 $ (in rad ≈ 143°)\n- $ 3x \approx 4.71 = 3\pi/2 $ → $ \sin = -1 $\n- $ 4x = 10 $ rad ≈ 573° ≡ 573 - 360×1 = 213° → $ \cos(213°) = \cos(180+33) = -\cos(33°) ≈ -0.836 $\n→ $ g ≈ -1 - 0.836 = -1.836 < -1 $", "So $ g(x) = -1 $ occurs, but we seek $ |g(x)| = 1 $", "In $ [0, \pi] $, $ g(x) = 1 $ at:\n- $ x = 0 $\n- $ x = \pi $", "But is there another point where $ g(x) = 1 $? Try $ x \approx 3.0 $:\n- $ 3x = 9 $ rad ≈ $ 9 - 2\pi \approx 9 - 6.28 = 2.72 $ rad ≈ 156° → $ \sin \approx 0.309 $\n- $ 4x = 12 $ rad ≈ $ 12 - 2\pi \ imes 1 = 12 - 6.28 = 5.72 $ rad ≈ 329° → $ \cos(329°) = \cos(-31°) ≈ 0.855 $ → $ g ≈ 1.164 > 1 $", "So $ g(x) > 1 $ briefly in $ (0, \pi) $ — but $ g(0) = 1 $, then increases, then decreases to $ g(\pi/3) = -0.5 $, then stays below — so only one time $ g(x) = 1 $ at endpoint.", "Similarly, since $ g(x) \ o 1 $ at $ x = \pi $, and no other peak reaches 1, likely only two points where $ |g(x)| = 1 $: $ x = 0 $ and $ x = \pi $", "But at $ x = 0 $ and $ x = \pi $, same value. Are they distinct in domain? Yes — $ t = 0 $ and $ t = 7 $, but $ t = 7 $ is not included, $ t \in [0,7) $, so $ x = 0 $ is included, $ x = \pi $ is included.", "But $ t = 7 \Rightarrow x = \pi $, so include.", "Now, does $ g(x) = -1 $ occur? Yes — e.g., near $ x = 2.4 $ as above, $ g \approx -1.2 $, so $ |g| = 1.2 <br/>\ne 1 $. But we need $ |g(x)| = 1 $", "Look for solutions to $ g(x) = \pm 1 $", "- $ g(x) = 1 $: occurs at $ x = 0 $ and $ x = \pi $ — two times\n- $ g(x) = -1 $: since $ g $ is continuous, and goes from negative values to 0 at $ 5\pi/6 $, then to 1 at $ \pi $, it must pass through $ -1 $ on the way up from $ \pi/3 $ to $ 5\pi/6 $? But at $ \pi/3 $, $ g = -0.5 $, and increases to 0 at $ 5\pi/6 $, so never reaches $ -1 $", "Wait—$ g(\pi/2) = 0 $, $ g(\pi) = 1 $, but minimum in between?", "Check derivative sign.", "But earlier at $ x = 2.5 $, $ g \approx -1.84 $, so yes, $ g(x) < -1 $ somewhere in $ (\pi/2, \pi) $", "When does it cross $ -1 $? Once from below? But $ g(\pi/2) = 0 $, $ g(\pi) = 1 $, so if it goes below, must cross $ -1 $", "Suppose $ g(x) = -1 $ has one solution in $ (\pi/2, \pi) $ near where it drops.", "But we also must check for $ g(x) = -1 $ in $ (0, \pi) $ — only one interval.", "Similarly, $ g(x) = 1 $: only at $ x = 0 $ and $ x = \pi $? But at $ x = 3.5 $, $ g \approx $?", "Estimate: $ 3x = 10.5 $ rad ≈ $ 10.5 - 3\pi \approx 10.5 - 9.42 = 1.08 $ rad → $ \sin \approx 0.87 $\n$ 4x = 14 $ rad ≈ $ 14 - 4\pi \approx 14 - 12.56 = 1.44 $ rad → $ \cos \approx \cos(82.5^\circ) \approx 0.13 $ → $ g \approx 1.0 $ — close", "But $ g(0) = 1 $, $ g(\pi) = 1 $, and derivative at 0:\n$ g'(0) = 3\cos(0) - 4\sin(0) = 3 > 0 $, so increases from 1.", "So it starts at 1, goes up slightly, then down — crosses $ y = 1 $ only at $ x = 0 $, but immediately exceeds it — so only one point where $ g(x) = 1 $ is at the endpoint? But $ x = 0 $ is included.", "Similarly, $ x = \pi $ is the only point where $ g(x) = 1 $ and derivative is zero? Not necessarily — could have flat?", "But since it starts at 1, $ x = 0 $ is one solution.", "Later, as $ g(x) $ decreases to negative values, then increases back to 1 at $ x = \pi $, it must cross $ y = 1 $ only at $ x = \pi $, but does not return to 1 before? From $ x = \pi - \epsilon $, $ g(\pi - \epsilon) < 1 $, since it drops.", "But $ g(x) = 1 $ only at $ x = 0 $ and $ x = \pi $"]