$0 \le \theta < 2\pi$ のとき、次の不等式を解く。 (1) $\cos(\theta + \frac{\pi}{6}) \ge \frac{1}{2}$ (2) $\sin(\theta - \frac{\pi}{4}) < \frac{\sqrt{3}}{2}$ (3) $\tan(\theta + \frac{\pi}{4}) > \sqrt{3}$

1. 問題の内容

$0 \le \theta < 2\pi$ のとき、次の不等式を解く。

(1) $\cos(\theta + \frac{\pi}{6}) \ge \frac{1}{2}$

(2) $\sin(\theta - \frac{\pi}{4}) < \frac{\sqrt{3}}{2}$

(3) $\tan(\theta + \frac{\pi}{4}) > \sqrt{3}$

2. 解き方の手順

(1) $\cos(\theta + \frac{\pi}{6}) \ge \frac{1}{2}$

$t = \theta + \frac{\pi}{6}$ とおく。$0 \le \theta < 2\pi$ より、$\frac{\pi}{6} \le t < 2\pi + \frac{\pi}{6}$

$\cos t \ge \frac{1}{2}$ を解くと、

$0 \le t \le \frac{\pi}{3}$, $\frac{5\pi}{3} \le t \le 2\pi$

これに $t = \theta + \frac{\pi}{6}$ を代入して、

$\frac{\pi}{6} \le \theta + \frac{\pi}{6} \le \frac{\pi}{3} \Rightarrow 0 \le \theta \le \frac{\pi}{6}$

$\frac{5\pi}{3} \le \theta + \frac{\pi}{6} < 2\pi + \frac{\pi}{6} \Rightarrow \frac{5\pi}{3} - \frac{\pi}{6} \le \theta < 2\pi$

$\Rightarrow \frac{10\pi - \pi}{6} \le \theta < 2\pi \Rightarrow \frac{3\pi}{2} \le \theta < 2\pi$

(2) $\sin(\theta - \frac{\pi}{4}) < \frac{\sqrt{3}}{2}$

$t = \theta - \frac{\pi}{4}$ とおく。$0 \le \theta < 2\pi$ より、$-\frac{\pi}{4} \le t < 2\pi - \frac{\pi}{4}$

$\sin t < \frac{\sqrt{3}}{2}$ を解くと、

$-\frac{\pi}{4} \le t < \frac{\pi}{3}$, $\frac{2\pi}{3} < t < 2\pi - \frac{\pi}{4}$

これに $t = \theta - \frac{\pi}{4}$ を代入して、

$-\frac{\pi}{4} \le \theta - \frac{\pi}{4} < \frac{\pi}{3} \Rightarrow 0 \le \theta < \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12}$

$\frac{2\pi}{3} < \theta - \frac{\pi}{4} < 2\pi - \frac{\pi}{4} \Rightarrow \frac{2\pi}{3} + \frac{\pi}{4} < \theta < 2\pi$

$\Rightarrow \frac{8\pi + 3\pi}{12} < \theta < 2\pi \Rightarrow \frac{11\pi}{12} < \theta < 2\pi$

(3) $\tan(\theta + \frac{\pi}{4}) > \sqrt{3}$

$t = \theta + \frac{\pi}{4}$ とおく。$0 \le \theta < 2\pi$ より、$\frac{\pi}{4} \le t < 2\pi + \frac{\pi}{4}$

$\tan t > \sqrt{3}$ を解くと、

$\frac{\pi}{3} < t < \frac{\pi}{2}$, $\frac{4\pi}{3} < t < \frac{3\pi}{2}$

これに $t = \theta + \frac{\pi}{4}$ を代入して、

$\frac{\pi}{3} < \theta + \frac{\pi}{4} < \frac{\pi}{2} \Rightarrow \frac{\pi}{3} - \frac{\pi}{4} < \theta < \frac{\pi}{2} - \frac{\pi}{4} \Rightarrow \frac{\pi}{12} < \theta < \frac{\pi}{4}$

$\frac{4\pi}{3} < \theta + \frac{\pi}{4} < \frac{3\pi}{2} \Rightarrow \frac{4\pi}{3} - \frac{\pi}{4} < \theta < \frac{3\pi}{2} - \frac{\pi}{4}$

$\Rightarrow \frac{16\pi - 3\pi}{12} < \theta < \frac{6\pi - \pi}{4} \Rightarrow \frac{13\pi}{12} < \theta < \frac{5\pi}{4}$

3. 最終的な答え

(1) $0 \le \theta \le \frac{\pi}{6}$, $\frac{3\pi}{2} \le \theta < 2\pi$

(2) $0 \le \theta < \frac{7\pi}{12}$, $\frac{11\pi}{12} < \theta < 2\pi$

(3) $\frac{\pi}{12} < \theta < \frac{\pi}{4}$, $\frac{13\pi}{12} < \theta < \frac{5\pi}{4}$