이학계열 1-3번

$\displaystyle \frac{1}{32}\leq x\leq1$ 이므로 $\displaystyle -5\leq\log_2x\leq0$ 즉, $\displaystyle \frac{-19\pi}{4}\leq\pi\log_2x+\frac\pi4\leq\frac\pi4$$\displaystyle \sin\left(\pi\log_2x+\frac\pi4\right)=-\frac{\sqrt2}{2}$$\displaystyle \pi\log_2x+\frac\pi4=\boxed{-\frac34\pi}$ 또는 $\displaystyle \pi\log_2x+\frac\pi4=\boxed{-\frac14\pi}$, $\displaystyle \pi\log_2x+\frac\pi4=\boxed{-\frac{11}{4}\pi}$ 또는 $\displaystyle \pi\log_2x+\frac\pi4=\boxed{-\frac94\pi}$$\displaystyle \pi\log_2x+\frac\pi4=\boxed{-\frac{19}{4}\pi}$ 또는 $\displaystyle \pi\log_2x+\frac\pi4=\boxed{-\frac{17}{4}\pi}$따라서 $\displaystyle x$는$\displaystyle x=\frac12$ 또는 $\displaystyle x=\frac{\sqrt2}{2}$$\displaystyle x=\frac18$ 또는 $\displaystyle x=\frac{\sqrt2}{8}$$\displaystyle x=\frac1{32}$ 또는 $\displaystyle x=\frac{\sqrt2}{32}$이므로답 $\displaystyle \frac{21}{32}(1+\sqrt2)$