$0 \le \theta < 2\pi$ のとき、次の方程式を満たす$\theta$の値を求めよ。 (1) $\sin\theta - \cos\theta = -\frac{1}{\sqrt{2}}$ (2) $\sqrt{3}\sin\theta + \cos\theta + 1 = 0$

1. 問題の内容

$0 \le \theta < 2\pi$ のとき、次の方程式を満たす$\theta$の値を求めよ。

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

(2) $\sqrt{3}\sin\theta + \cos\theta + 1 = 0$

2. 解き方の手順

(1)

$\sin\theta - \cos\theta = -\frac{1}{\sqrt{2}}$

左辺を合成すると、

$\sqrt{2}\sin(\theta - \frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$

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

$0 \le \theta < 2\pi$ より、$-\frac{\pi}{4} \le \theta - \frac{\pi}{4} < \frac{7\pi}{4}$

$\theta - \frac{\pi}{4} = \frac{7\pi}{6}, \frac{11\pi}{6}$

$\theta = \frac{7\pi}{6} + \frac{\pi}{4}, \frac{11\pi}{6} + \frac{\pi}{4}$

$\theta = \frac{14\pi + 3\pi}{12}, \frac{22\pi + 3\pi}{12}$

$\theta = \frac{17\pi}{12}, \frac{25\pi}{12}$

(2)

$\sqrt{3}\sin\theta + \cos\theta + 1 = 0$

$\sqrt{3}\sin\theta + \cos\theta = -1$

左辺を合成すると、

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

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

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

$\theta + \frac{\pi}{6} = \frac{7\pi}{6}, \frac{11\pi}{6}$

$\theta = \frac{7\pi}{6} - \frac{\pi}{6}, \frac{11\pi}{6} - \frac{\pi}{6}$

$\theta = \frac{6\pi}{6}, \frac{10\pi}{6}$

$\theta = \pi, \frac{5\pi}{3}$

3. 最終的な答え

(1) $\theta = \frac{17\pi}{12}, \frac{25\pi}{12}$

(2) $\theta = \pi, \frac{5\pi}{3}$