合成して解く三角方程式・不等式（B）

(1) $2\sin\left(\theta-\dfrac{\pi}{6}\right)=\sqrt{2}$ より $\sin\left(\theta-\dfrac{\pi}{6}\right)=\dfrac{1}{\sqrt{2}}$． $t=\theta-\dfrac{\pi}{6}$ とおくと $-\dfrac{\pi}{6}\leqq t<\dfrac{11}{6}\pi$ で，$t=\dfrac{\pi}{4},\ \dfrac{3}{4}\pi$．よって $\theta=\dfrac{5}{12}\pi,\ \dfrac{11}{12}\pi$ (2) $\sqrt{2}\sin\left(\theta+\dfrac{\pi}{4}\right)<1$ より $\sin\left(\theta+\dfrac{\pi}{4}\right)<\dfrac{1}{\sqrt{2}}$． $t=\theta+\dfrac{\pi}{4}$ とおくと $\dfrac{\pi}{4}\leqq t<\dfrac{9}{4}\pi$ で，$\dfrac{3}{4}\pi<t<\dfrac{9}{4}\pi$．よって $\dfrac{\pi}{2}<\theta<2\pi$