2. Zadatak

Uz uvjet $\sin2x>0$ iz $\log_8(\sin2x)=-\dfrac13=\log_8 8^{-\frac13}=\log_8\dfrac12$ dobivamo

$$\sin2x=\frac12,$$

odakle je zadovoljen uvjet i imamo

$$2x=\frac\pi6+2k\pi\quad\text{ili}\quad2x=\frac{5\pi}6+2k\pi,$$

tj.

$$x=\frac\pi{12}+k\pi\quad\text{ili}\quad x=\frac{5\pi}{12}+k\pi.$$

Na intervalu $[0,2\pi]$ rješenja su $x_1=\dfrac\pi{12}$, $x_2=\dfrac{5\pi}{12}$, $x_3=\dfrac{13\pi}{12}$, $x_1=\dfrac{17\pi}{12}$.

Zbroj svih rješenja je $\dfrac\pi{12}+\dfrac{5\pi}{12}+\dfrac{13\pi}{12}+\dfrac{17\pi}{12}=3\pi$.