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Home » $ k = \frac\ln(2)4 \approx \frac0.6934 = 0.173 $ per year - Nelissen Grade advocaten

$ k = \frac\ln(2)4 \approx \frac0.6934 = 0.173 $ per year - Nelissen Grade advocaten

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 The general solution for $ \cos \phi = \frac{\sqrt{3}}{2} $ is:

📌 \phi = 30^\circ + 360^\circ k \quad \text{or} \quad \phi = 330^\circ + 360^\circ k, \quad k \in \mathbb{Z}

📌 Since $ \phi = 3\theta $, we substitute:

📌 3\theta = 30^\circ + 360^\circ k \quad \Rightarrow \quad \theta = 10^\circ + 120^\circ k

📌 3\theta = 330^\circ + 360^\circ k \quad \Rightarrow \quad \theta = 110^\circ + 120^\circ k

📌 We now find all $ \theta \in [0^\circ, 360^\circ] $ by plugging integer values of $ k $.

📌 For $ \theta = 10^\circ + 120^\circ k $:

📌 $ k = 0 $: $ 10^\circ $

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f(x) = \sec^2 x + \csc^2 x + (\cos^2 x + \sin^2 x) + 2 - 2 = \sec^2 x...
Now express in terms of sine and cosine:
f(x) = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} + 1 = \frac{\sin^2 x +...
Using the identity $ \sin(2x) = 2\sin x \cos x $, we have:
\sin x \cos x = \frac{1}{2} \sin 2x \Rightarrow \sin^2 x \cos^2 x = \f...
f(x) = \frac{1}{\frac{1}{4} \sin^2 2x} + 1 = \frac{4}{\sin^2 2x} + 1
Since $ 0 < x < \frac{\pi}{2} $, then $ 0 < 2x < \pi $, and $ \sin 2x...
f\left(\frac{\pi}{4}\right) = \frac{4}{1} + 1 = 5
Thus, the minimum value is $ \boxed{5} $.
Question: A mammalogist observes that the social cohesion index $ S $...

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