11.3. Find the roots of the equation: $\left(x^{3}-2\right)\left(2^{\sin x}-1\right)+\left(2^{x^{3}}-4\right) \sin x=0$.
Answer. $\sqrt[3]{2}, \pi n(n \in \mathbf{Z})$.

Solution. From the fact that the function $\mathrm{y}=2^{\mathrm{t}}$ is increasing, it follows:

1) if $\sin x>0$, then $2^{\sin x}-1>0$; if $\sin x0$, then $2^{x^{3}}-4>0$; if $x^{3}-20$, then $\left(2^{x^{3}}-4\right) \cdot \sin x>0$; if $\left(x^{3}-2\right)\left(2^{\sin x}-1\right)<0$, then $\left(2^{x^{3}}-4\right) \cdot \sin x<0$; that is, the signs of the expressions $\left(x^{3}-2\right)\left(2^{\sin x}-1\right)$ and $\left(2^{x^{3}}-4\right) \cdot \sin x$ coincide.

Therefore, each term on the left side of the equation must be zero, that is, $x^{3}=2$ or $\sin x=0$.