2. Zadatak

Vrijedi $(f\circ g)(x)=f(g(x))=4^{8^{-2^{g(x)}}}$. Zato je

$$4^{8^{-2^{g(x)}}}=2^{4^{4^{\sin x}}}.$$

Slijedi redom

$$\begin{aligned} 2^{2\cdot8^{-2^{g(x)}}}&=2^{4^{4^{\sin x}}}\\ 2\cdot8^{-2^{g(x)}}&=4^{4^{\sin x}}\\ 2\cdot2^{-3\cdot2^{g(x)}}&=2^{2\cdot4^{\sin x}}\\ 2^{1-3\cdot2^{g(x)}}&=2^{2\cdot4^{\sin x}}\\ 1-3\cdot2^{g(x)}&=2\cdot4^{\sin x}\\ 1-3\cdot2^{g(x)}&=2\cdot2^{2\sin x}\\ 1-3\cdot2^{g(x)}&=2^{1+2\sin x}\\ 3\cdot2^{g(x)}&=1-2^{1+2\sin x}\\ 2^{g(x)}&=\frac13\left(1-2^{1+2\sin x}\right)\\ g(x)&=\log_2\frac{1-2^{1+2\sin x}}3. \end{aligned}$$

Odredimo domenu funkcije $g$.

Iz uvjeta $\dfrac{1-2^{1+2\sin x}}3>0$ redom slijedi $2^{1+2\sin x}<1$, $1+2\sin x<0$, $\sin x<-\dfrac12$.

Konačno,

$$x\in\bigcup_{k\in\mathbb Z} \left\langle\frac{7\pi}{6}+2k\pi,\frac{11\pi}{6}+2k\pi\right\rangle.$$