$0 \leq \theta < 2\pi$ のとき、以下の三角関数の方程式と不等式を解きます。 (1) $\sin(\theta - \frac{\pi}{3}) = -\frac{1}{2}$ (2) $\cos(2\theta + \frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ (3) $\tan(\theta - \frac{\pi}{6}) > 1$ (4) $\sin(2\theta + \frac{\pi}{6}) \leq -\frac{1}{2}$

はい、承知いたしました。画像に書かれている三角関数の方程式と不等式を解きます。

1. 問題の内容

$0 \leq \theta < 2\pi$ のとき、以下の三角関数の方程式と不等式を解きます。

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

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

(3) $\tan(\theta - \frac{\pi}{6}) > 1$

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

2. 解き方の手順

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

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

$t = \frac{7\pi}{6}, \frac{11\pi}{6}$

$\theta - \frac{\pi}{3} = \frac{7\pi}{6} \Rightarrow \theta = \frac{7\pi}{6} + \frac{\pi}{3} = \frac{9\pi}{6} = \frac{3\pi}{2}$

$\theta - \frac{\pi}{3} = \frac{11\pi}{6} \Rightarrow \theta = \frac{11\pi}{6} + \frac{\pi}{3} = \frac{13\pi}{6} = \frac{13\pi}{6} - 2\pi = \frac{\pi}{6} + \frac{\pi}{3} = \frac{13\pi}{6}$

ただし、$0 \leq \theta < 2\pi$ より、$\theta = \frac{3\pi}{2}, \frac{13\pi}{6} - \frac{12\pi}{6} = \frac{\pi}{6}$。

$t$ の範囲を考えると、$\theta$ は $0 \leq \theta < 2\pi$ なので、 $-\frac{\pi}{3} \leq t < 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$。

よって、$t = \frac{7\pi}{6}, \frac{11\pi}{6}$ は正しいので、$\theta = \frac{3\pi}{2}, \frac{13\pi}{6}$ は正しい。

$\theta = \frac{3\pi}{2}, \frac{11\pi}{6} + \frac{2\pi}{6} = \frac{13\pi}{6} = 2\pi + \frac{\pi}{6}$ であり、$0 \le \theta < 2\pi$ の範囲より$\theta = \frac{3\pi}{2}, \frac{11\pi}{6} + \frac{2\pi}{6} = \frac{3\pi}{2}$ と $\frac{13\pi}{6}-2\pi = \frac{\pi}{6} \pi$ なので $\frac{11\pi}{6} \Rightarrow \theta = \frac{11\pi}{6} + \frac{2\pi}{6} = \frac{13\pi}{6}$ が必要で、$\frac{13\pi}{6} - 2\pi = \frac{\pi}{6}$ なので $\theta = \frac{3\pi}{2}, \frac{11\pi}{6}+\frac{\pi}{6}2\pi$なので

$\frac{\pi}{6}$ は $\frac{13\pi}{6}$ と $\theta = \frac{\pi}{6}, \frac{3\pi}{2}$

$\theta - \frac{\pi}{3} = \frac{7\pi}{6}, \frac{11\pi}{6}$なので $\theta = \frac{3\pi}{2}, \frac{13\pi}{6} > 2\pi$

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

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

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

$2\theta + \frac{\pi}{3} = \frac{\pi}{6} \Rightarrow 2\theta = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6} \Rightarrow \theta = -\frac{\pi}{12}$

$2\theta + \frac{\pi}{3} = \frac{11\pi}{6} \Rightarrow 2\theta = \frac{11\pi}{6} - \frac{\pi}{3} = \frac{9\pi}{6} = \frac{3\pi}{2} \Rightarrow \theta = \frac{3\pi}{4}$

$\theta$の範囲 $0 \leq \theta < 2\pi$を考えると、

$\theta = -\frac{\pi}{12} + \pi = \frac{11\pi}{12}$

$\theta = -\frac{\pi}{12} + 2\pi = \frac{23\pi}{12}$

$\theta = \frac{3\pi}{4}$

$\theta = \frac{3\pi}{4} + \pi = \frac{7\pi}{4}$

$t$ の範囲を考えると $\frac{\pi}{3} \le t < 4\pi + \frac{\pi}{3}$

$t = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{13\pi}{6}, \frac{23\pi}{6}$ なので、$2\theta = \frac{11\pi}{6}, \frac{3\pi}{2}, \frac{23\pi}{6}$

よって、 $\theta = -\frac{\pi}{12}+ \pi = \frac{11\pi}{12} \theta = \frac{3\pi}{4}$

よって $\theta = \frac{11\pi}{12}, \frac{3\pi}{4}, \frac{23\pi}{12}$

(3) $\tan(\theta - \frac{\pi}{6}) > 1$

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

$\frac{\pi}{4} < t < \frac{\pi}{2}, \frac{5\pi}{4} < t < \frac{3\pi}{2}$

$\frac{\pi}{4} < \theta - \frac{\pi}{6} < \frac{\pi}{2} \Rightarrow \frac{\pi}{4} + \frac{\pi}{6} < \theta < \frac{\pi}{2} + \frac{\pi}{6} \Rightarrow \frac{5\pi}{12} < \theta < \frac{2\pi}{3}$

$\frac{5\pi}{4} < \theta - \frac{\pi}{6} < \frac{3\pi}{2} \Rightarrow \frac{5\pi}{4} + \frac{\pi}{6} < \theta < \frac{3\pi}{2} + \frac{\pi}{6} \Rightarrow \frac{17\pi}{12} < \theta < \frac{5\pi}{3}$

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

$2\theta + \frac{\pi}{6} = t$ とおくと、$\sin t \leq -\frac{1}{2}$

$\frac{7\pi}{6} \leq t \leq \frac{11\pi}{6}$

$\frac{7\pi}{6} \leq 2\theta + \frac{\pi}{6} \leq \frac{11\pi}{6} \Rightarrow \frac{6\pi}{6} \leq 2\theta \leq \frac{10\pi}{6} \Rightarrow \pi \leq 2\theta \leq \frac{5\pi}{3} \Rightarrow \frac{\pi}{2} \leq \theta \leq \frac{5\pi}{6}$

$\frac{7\pi}{6} + 2\pi \leq 2\theta + \frac{\pi}{6} \leq \frac{11\pi}{6} + 2\pi \Rightarrow \frac{19\pi}{6} \leq 2\theta + \frac{\pi}{6} \leq \frac{23\pi}{6} \Rightarrow \frac{18\pi}{6} \leq 2\theta \leq \frac{22\pi}{6} \Rightarrow 3\pi \leq 2\theta \leq \frac{11\pi}{3} \Rightarrow \frac{3\pi}{2} \leq \theta \leq \frac{11\pi}{6} \Rightarrow 3\pi \leq 2\theta \leq \frac{11\pi}{3}$

$\frac{\pi}{2} \leq \theta \leq \frac{5\pi}{6}$または$\frac{3\pi}{2} \leq \theta \leq \frac{11\pi}{6}$

3. 最終的な答え

(1) $\theta = \frac{3\pi}{2}, \frac{7\pi}{6}$

(2) $\theta = \frac{3\pi}{4}, \frac{11\pi}{12}, \frac{19\pi}{12}, \frac{23\pi}{12}$

(3) $\frac{5\pi}{12} < \theta < \frac{2\pi}{3}$, $\frac{17\pi}{12} < \theta < \frac{5\pi}{3}$

(4) $\frac{\pi}{2} \leq \theta \leq \frac{5\pi}{6}$, $\frac{3\pi}{2} \leq \theta \leq \frac{11\pi}{6}$