151116 初版 151116 更新

\(\sin\left(2\theta+\dfrac{\pi}{4}\right)=\dfrac{1}{2}\) を満たす θ
一般には,
\(2\theta+\dfrac{\pi}{4} = \dfrac{1+12n}{6}\pi\),  \(\dfrac{5+12n}{6}\pi\)   (n は整数)
すなわち,
\(\theta = \dfrac{-1+24n}{24}\pi\),  \(\dfrac{7+24n}{24}\pi\)   (n は整数)
0 ≦ θ < 2π では,
\(\theta = \dfrac{7}{24}\pi\),   \(\dfrac{23}{24}\pi\),   \(\dfrac{31}{24}\pi\),   \(\dfrac{47}{24}\pi\)

\(\sin\left(2\theta+\dfrac{\pi}{4}\right) >\dfrac{1}{2}\) を満たす θ
一般には,
\(\dfrac{1+12n}{6}\pi < 2\theta+\dfrac{\pi}{4}\) \(< \dfrac{5+12n}{6}\pi\)   (n は整数)
すなわち,
\(\dfrac{-1+24n}{24}\pi < \theta\) \(< \dfrac{7+24n}{24}\pi\)   (n は整数)
0 ≦ θ < 2π では,
\(0\leqq \theta < \dfrac{7}{24}\pi\),  \(\dfrac{23}{24}\pi < \theta < \dfrac{31}{24}\pi\),   \(\dfrac{47}{24}\pi < \theta < 2\pi\),  

\(\sin\left(2\theta+\dfrac{\pi}{4}\right) <\dfrac{1}{2}\) を満たす θ
一般には,
\(\dfrac{5+12n}{6}\pi < 2\theta+\dfrac{\pi}{4}\) \(< \dfrac{13+12n}{6}\pi\)   (n は整数)
すなわち,
\(\dfrac{7+24n}{24}\pi < \theta\) \(< \dfrac{23+24n}{24}\pi\)   (n は整数)
0 ≦ θ < 2π では,
\(\dfrac{7}{24}\pi < \theta\) \(< \dfrac{23}{24}\pi\),   \(\dfrac{31}{24}\pi < \theta\) \(< \dfrac{47}{24}\pi\)

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