151116 初版 151116 更新
例
\(\tan\left(\theta+\dfrac{\pi}{3}\right)=1\) を満たす θ
一般には，
\(\theta+\dfrac{\pi}{3} = \dfrac{\pi}{4}+n\pi\)
(n は整数)
すなわち，
\(\theta = \dfrac{-1+12n}{12}\pi\)
(n は整数)
0 ≦ θ < 2π では，
\(\theta = \dfrac{11}{12}\pi\),
\(\dfrac{23}{12}\pi\)
例
\(\tan\left(\theta+\dfrac{\pi}{3}\right) > 1\) を満たす θ
一般には，
\(\dfrac{\pi}{4}+n\pi < \theta+\dfrac{\pi}{3}\)
\(< \dfrac{\pi}{2}+n\pi\)
(n は整数)
すなわち，
\(\dfrac{-1+12n}{12}\pi < \theta\)
\(< \dfrac{1+6n}{6}\pi\)
(n は整数)
0 ≦ θ < 2π では，
\(0\leqq \theta < \dfrac{\pi}{6}\),
\(\dfrac{11}{12}\pi < \theta < \dfrac{7}{6}\pi\),
\(\dfrac{23}{12}\pi < \theta < 2\pi\)
例
\(\tan\left(\theta+\dfrac{\pi}{3}\right) < 1\) を満たす θ
一般には，
\(-\dfrac{\pi}{2}+n\pi < \theta+\dfrac{\pi}{3}\)
\(< \dfrac{\pi}{4}+n\pi\)
(n は整数)
すなわち，
\(\dfrac{-5+6n}{6}\pi < \theta\)
\(< \dfrac{-1+12n}{12}\pi\)
(n は整数)
0 ≦ θ < 2π では，
\(\dfrac{\pi}{6} < \theta < \dfrac{11}{12}\pi\),
\(\dfrac{7}{6}\pi < \theta < \dfrac{23}{12}\pi\)