$0 \leq \theta < 2\pi$ の範囲で、次の方程式を解きます。 $\sin \theta + \cos \theta = -1$

## (1) の問題

1. 問題の内容

$0 \leq \theta < 2\pi$ の範囲で、次の方程式を解きます。

$\sin \theta + \cos \theta = -1$

2. 解き方の手順

$\sin \theta + \cos \theta = -1$ の両辺を$\sqrt{2}$で割ると、

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

$\cos \frac{\pi}{4} \sin \theta + \sin \frac{\pi}{4} \cos \theta = -\frac{1}{\sqrt{2}}$

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

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

$\theta + \frac{\pi}{4} = \frac{5}{4}\pi, \frac{7}{4}\pi$

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

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

3. 最終的な答え

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

## (2) の問題

1. 問題の内容

$0 \leq \theta < 2\pi$ の範囲で、次の方程式を解きます。

$\sqrt{3} \sin \theta - \cos \theta - \sqrt{2} = 0$

2. 解き方の手順

$\sqrt{3} \sin \theta - \cos \theta = \sqrt{2}$

両辺を$2$で割ると、

$\frac{\sqrt{3}}{2} \sin \theta - \frac{1}{2} \cos \theta = \frac{\sqrt{2}}{2}$

$\cos \frac{\pi}{6} \sin \theta - \sin \frac{\pi}{6} \cos \theta = \frac{\sqrt{2}}{2}$

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

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

$\theta - \frac{\pi}{6} = \frac{\pi}{4}, \frac{3}{4}\pi$

$\theta = \frac{\pi}{4} + \frac{\pi}{6}, \frac{3}{4}\pi + \frac{\pi}{6}$

$\theta = \frac{3\pi}{12} + \frac{2\pi}{12}, \frac{9\pi}{12} + \frac{2\pi}{12}$

$\theta = \frac{5}{12}\pi, \frac{11}{12}\pi$

3. 最終的な答え

$\theta = \frac{5}{12}\pi, \frac{11}{12}\pi$