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

1. 問題の内容

$0 \le \theta < 2\pi$ のとき、次の方程式を解く。

(1) $2\cos^2{\theta} - 3\cos{\theta} + 1 = 0$

(2) $2\cos^2{\theta} - \sin{\theta} - 1 = 0$

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

2. 解き方の手順

(1) $2\cos^2{\theta} - 3\cos{\theta} + 1 = 0$

$\cos{\theta} = t$ とおくと、$2t^2 - 3t + 1 = 0$

$(2t - 1)(t - 1) = 0$

$t = \frac{1}{2}, 1$

したがって、$\cos{\theta} = \frac{1}{2}, 1$

$0 \le \theta < 2\pi$ より、$\theta = \frac{\pi}{3}, \frac{5\pi}{3}, 0$

(2) $2\cos^2{\theta} - \sin{\theta} - 1 = 0$

$2(1 - \sin^2{\theta}) - \sin{\theta} - 1 = 0$

$2 - 2\sin^2{\theta} - \sin{\theta} - 1 = 0$

$-2\sin^2{\theta} - \sin{\theta} + 1 = 0$

$2\sin^2{\theta} + \sin{\theta} - 1 = 0$

$\sin{\theta} = t$ とおくと、$2t^2 + t - 1 = 0$

$(2t - 1)(t + 1) = 0$

$t = \frac{1}{2}, -1$

したがって、$\sin{\theta} = \frac{1}{2}, -1$

$0 \le \theta < 2\pi$ より、$\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$

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

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

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

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

$\cos{\theta} = 0$ より $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$

$\sin{\theta} = \frac{\sqrt{3}}{2}$ より $\theta = \frac{\pi}{3}, \frac{2\pi}{3}$

したがって、$\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{\pi}{3}, \frac{2\pi}{3}$

3. 最終的な答え

(1) $\theta = 0, \frac{\pi}{3}, \frac{5\pi}{3}$

(2) $\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}$

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