$0 \le \theta < 2\pi$ のとき、次の方程式を解く。 (1) $\sin(\theta - \frac{\pi}{6}) = -\frac{1}{\sqrt{2}}$ (2) $\cos(\theta + \frac{\pi}{4}) = \frac{\sqrt{3}}{2}$

1. 問題の内容

$0 \le \theta < 2\pi$ のとき、次の方程式を解く。

(1) $\sin(\theta - \frac{\pi}{6}) = -\frac{1}{\sqrt{2}}$

(2) $\cos(\theta + \frac{\pi}{4}) = \frac{\sqrt{3}}{2}$

2. 解き方の手順

(1) $\sin(\theta - \frac{\pi}{6}) = -\frac{1}{\sqrt{2}}$

$\theta - \frac{\pi}{6} = t$ とおく。

$\sin t = -\frac{1}{\sqrt{2}}$

$0 \le \theta < 2\pi$ より、 $-\frac{\pi}{6} \le \theta - \frac{\pi}{6} < 2\pi - \frac{\pi}{6}$

$-\frac{\pi}{6} \le t < \frac{11\pi}{6}$

$\sin t = -\frac{1}{\sqrt{2}}$ を満たす $t$ は、

$t = \frac{5\pi}{4}, \frac{7\pi}{4}$

$\theta - \frac{\pi}{6} = \frac{5\pi}{4}$ より、 $\theta = \frac{5\pi}{4} + \frac{\pi}{6} = \frac{15\pi}{12} + \frac{2\pi}{12} = \frac{17\pi}{12}$

$\theta - \frac{\pi}{6} = \frac{7\pi}{4}$ より、 $\theta = \frac{7\pi}{4} + \frac{\pi}{6} = \frac{21\pi}{12} + \frac{2\pi}{12} = \frac{23\pi}{12}$

(2) $\cos(\theta + \frac{\pi}{4}) = \frac{\sqrt{3}}{2}$

$\theta + \frac{\pi}{4} = t$ とおく。

$\cos t = \frac{\sqrt{3}}{2}$

$0 \le \theta < 2\pi$ より、 $\frac{\pi}{4} \le \theta + \frac{\pi}{4} < 2\pi + \frac{\pi}{4}$

$\frac{\pi}{4} \le t < \frac{9\pi}{4}$

$\cos t = \frac{\sqrt{3}}{2}$ を満たす $t$ は、

$t = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{13\pi}{6}$

$\theta + \frac{\pi}{4} = \frac{\pi}{6}$ より、 $\theta = \frac{\pi}{6} - \frac{\pi}{4} = \frac{2\pi}{12} - \frac{3\pi}{12} = -\frac{\pi}{12}$

これは、$0 \le \theta < 2\pi$ を満たさない。

$\theta + \frac{\pi}{4} = \frac{11\pi}{6}$ より、 $\theta = \frac{11\pi}{6} - \frac{\pi}{4} = \frac{22\pi}{12} - \frac{3\pi}{12} = \frac{19\pi}{12}$

$\theta + \frac{\pi}{4} = \frac{13\pi}{6}$ より、 $\theta = \frac{13\pi}{6} - \frac{\pi}{4} = \frac{26\pi}{12} - \frac{3\pi}{12} = \frac{23\pi}{12}$

3. 最終的な答え

(1) $\theta = \frac{17\pi}{12}, \frac{23\pi}{12}$

(2) $\theta = \frac{19\pi}{12}, \frac{23\pi}{12}$