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

1. 問題の内容

$0 \le \theta < 2\pi$ のとき、次の方程式を解く。

(1) $\sin(\theta + \frac{\pi}{6}) = \frac{\sqrt{3}}{2}$

(2) $\cos(\theta - \frac{\pi}{3}) = -\frac{1}{\sqrt{2}}$

(3) $\sin(\theta + \frac{\pi}{4}) = \frac{1}{2}$

(4) $\tan(\theta + \frac{\pi}{6}) = 1$

2. 解き方の手順

(1) $\sin(\theta + \frac{\pi}{6}) = \frac{\sqrt{3}}{2}$

$\theta + \frac{\pi}{6} = t$ と置くと、$\sin t = \frac{\sqrt{3}}{2}$。

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

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

$t = \frac{\pi}{3}, \frac{2\pi}{3}$。

したがって、

$\theta + \frac{\pi}{6} = \frac{\pi}{3}$ より、$\theta = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$。

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

(2) $\cos(\theta - \frac{\pi}{3}) = -\frac{1}{\sqrt{2}}$

$\theta - \frac{\pi}{3} = t$ と置くと、$\cos t = -\frac{1}{\sqrt{2}}$。

$0 \le \theta < 2\pi$ より、$-\frac{\pi}{3} \le t < 2\pi - \frac{\pi}{3}$。

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

$t = \frac{3\pi}{4}, \frac{5\pi}{4}$。

したがって、

$\theta - \frac{\pi}{3} = \frac{3\pi}{4}$ より、$\theta = \frac{3\pi}{4} + \frac{\pi}{3} = \frac{9\pi}{12} + \frac{4\pi}{12} = \frac{13\pi}{12}$。

$\theta - \frac{\pi}{3} = \frac{5\pi}{4}$ より、$\theta = \frac{5\pi}{4} + \frac{\pi}{3} = \frac{15\pi}{12} + \frac{4\pi}{12} = \frac{19\pi}{12}$。

(3) $\sin(\theta + \frac{\pi}{4}) = \frac{1}{2}$

$\theta + \frac{\pi}{4} = t$ と置くと、$\sin t = \frac{1}{2}$。

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

$\sin t = \frac{1}{2}$ を満たす $t$ は、

$t = \frac{\pi}{6}, \frac{5\pi}{6}$。

したがって、

$\theta + \frac{\pi}{4} = \frac{\pi}{6}$ より、$\theta = \frac{\pi}{6} - \frac{\pi}{4} = \frac{2\pi}{12} - \frac{3\pi}{12} = -\frac{\pi}{12}$。これは範囲外なので、$2\pi$ を足すと $\theta = 2\pi - \frac{\pi}{12} = \frac{23\pi}{12}$。

$\theta + \frac{\pi}{4} = \frac{5\pi}{6}$ より、$\theta = \frac{5\pi}{6} - \frac{\pi}{4} = \frac{10\pi}{12} - \frac{3\pi}{12} = \frac{7\pi}{12}$。

(4) $\tan(\theta + \frac{\pi}{6}) = 1$

$\theta + \frac{\pi}{6} = t$ と置くと、$\tan t = 1$。

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

$\tan t = 1$ を満たす $t$ は、

$t = \frac{\pi}{4}, \frac{5\pi}{4}$。

したがって、

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

$\theta + \frac{\pi}{6} = \frac{5\pi}{4}$ より、$\theta = \frac{5\pi}{4} - \frac{\pi}{6} = \frac{15\pi}{12} - \frac{2\pi}{12} = \frac{13\pi}{12}$。

3. 最終的な答え

(1) $\theta = \frac{\pi}{6}, \frac{\pi}{2}$

(2) $\theta = \frac{13\pi}{12}, \frac{19\pi}{12}$

(3) $\theta = \frac{7\pi}{12}, \frac{23\pi}{12}$

(4) $\theta = \frac{\pi}{12}, \frac{13\pi}{12}$