三角関数

これを

左辺を因数分解すると， $$ \begin{aligned} (\sqrt3 \sin\theta + \cos\theta)(\sin\theta + \cos\theta)=0 \end{aligned} $$ よって， $$ \begin{aligned} \sqrt3 \sin\theta + \cos\theta = 0　または　\sin\theta + \cos\theta = 0 \end{aligned} $$ （ア）$\sqrt3 \sin\theta + \cos\theta = 0$　のとき $$ \begin{aligned} \sqrt3 \sin\theta + \cos\theta &= 0 \\ \Leftrightarrow 2(\frac{\sqrt3}{2}\sin\theta + \frac{1}{2} \cos\theta) &= 0\\ \Leftrightarrow \frac{\sqrt3}{2}\sin\theta + \frac{1}{2} \cos\theta &= 0\\ \Leftrightarrow \sin(\theta + \frac{\pi}{6}) &= 0 \end{aligned} $$ 　$0 \leq \theta < 2\pi$　より $\frac{\pi}{6} \leq \theta+ \frac{\pi}{6} < \frac{13\pi}{6}$　だから $$ \begin{aligned} \theta + \frac{\pi}{6} &= \pi,2\pi\\ \Leftrightarrow \theta &= \frac{5\pi}{6},\frac{11\pi}{6} \end{aligned} $$ （イ）$\sin\theta + \cos\theta = 0$　のとき\\ ・ ・ ・