$0 \le \theta < 2\pi$ の範囲で、以下の三角関数の方程式を解く問題です。 (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

$0 \le \theta < 2\pi$ の範囲で、以下の三角関数の方程式を解く問題です。

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

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

$0 \le \theta < 2\pi$ より、$\frac{\pi}{6} \le t < \frac{13\pi}{6}$。

$\sin t = \frac{\sqrt{3}}{2}$ となる $t$ は、

$t = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3}$。

$\theta = t - \frac{\pi}{6}$ なので、

$\theta = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$,

$\theta = \frac{2\pi}{3} - \frac{\pi}{6} = \frac{\pi}{2}$,

$\theta = \frac{7\pi}{3} - \frac{\pi}{6} = \frac{13\pi}{6}$,

$\theta = \frac{8\pi}{3} - \frac{\pi}{6} = \frac{5\pi}{2}$.

ただし、$\theta < 2\pi$ であるから、

$\theta = \frac{\pi}{6}, \frac{\pi}{2}$

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

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

$0 \le \theta < 2\pi$ より、$-\frac{\pi}{3} \le t < \frac{5\pi}{3}$。

$\cos t = -\frac{1}{\sqrt{2}}$ となる $t$ は、

$t = \frac{3\pi}{4}, \frac{5\pi}{4}$。

$\theta = t + \frac{\pi}{3}$ なので、

$\theta = \frac{3\pi}{4} + \frac{\pi}{3} = \frac{13\pi}{12}$,

$\theta = \frac{5\pi}{4} + \frac{\pi}{3} = \frac{19\pi}{12}$.

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

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

$0 \le \theta < 2\pi$ より、$\frac{\pi}{4} \le t < \frac{9\pi}{4}$。

$\sin t = \frac{1}{2}$ となる $t$ は、

$t = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}$。

$\theta = t - \frac{\pi}{4}$ なので、

$\theta = \frac{\pi}{6} - \frac{\pi}{4} = -\frac{\pi}{12}$,

$\theta = \frac{5\pi}{6} - \frac{\pi}{4} = \frac{7\pi}{12}$,

$\theta = \frac{13\pi}{6} - \frac{\pi}{4} = \frac{23\pi}{12}$,

$\theta = \frac{17\pi}{6} - \frac{\pi}{4} = \frac{24\pi}{12} = \frac{31\pi}{12}$。

ただし、$0 \le \theta < 2\pi$ であるから、

$\theta = \frac{7\pi}{12}, \frac{23\pi}{12}$.

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

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

$0 \le \theta < 2\pi$ より、$\frac{\pi}{6} \le t < \frac{13\pi}{6}$。

$\tan t = 1$ となる $t$ は、

$t = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}$

$\theta = t - \frac{\pi}{6}$ なので、

$\theta = \frac{\pi}{4} - \frac{\pi}{6} = \frac{\pi}{12}$,

$\theta = \frac{5\pi}{4} - \frac{\pi}{6} = \frac{13\pi}{12}$,

$\theta = \frac{9\pi}{4} - \frac{\pi}{6} = \frac{25\pi}{12}$,

$\theta = \frac{13\pi}{4} - \frac{\pi}{6} = \frac{37\pi}{12}$.

ただし、$0 \le \theta < 2\pi$ であるから、

$\theta = \frac{\pi}{12}, \frac{13\pi}{12}$.

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