The equivalent of the given problem is that the difference - denoted by $K$ - between the left and right sides is positive for which $x$ values. Express the second and third terms of $K$ also in terms of $\sin x$. Since $\sin 3 x=\sin x\left(3-4 \sin ^{2} x\right)$,

$$\begin{gathered} \sin ^{2} 2 x=4 \sin ^{2} x \cos ^{2} x=4 \sin ^{2} x\left(1-\sin ^{2} x\right) \\ \sin ^{2} 3 x=\sin ^{2} x\left(9-24 \sin ^{2} x+16 \sin ^{4} x\right), \text { and thus } \\ K=\sin ^{2} x+\sin ^{2} 2 x-\sin ^{2} 3 x=\sin ^{2} x\left(-4+20 \sin ^{2} x-16 \sin ^{4} x\right)= \\ =-4 \sin ^{2} x\left(4 \sin ^{4} x-5 \sin ^{2} x+1\right)=-4 \sin ^{2} x\left(\sin ^{2} x-1\right)\left(4 \sin ^{2} x-1\right)= \\ =4 \sin ^{2} x \cos ^{2} x\left(4 \sin ^{2} x-1\right) \end{gathered}$$

(We factored the quadratic polynomial in $\sin ^{2} x$ using the fact that the roots of the equation $4 z^{2}-5 z+1=0$ are $z_{1}=1, z_{2}=1 / 4$.) The first two factors of the last form are never negative, but can be 0, specifically when - in the $\left(0^{\circ}, 360^{\circ}\right)$ interval - $x=0^{\circ}, 180^{\circ}$, or $90^{\circ}, 270^{\circ}$. After excluding these $x$ values, $K$ is positive if and only if the third factor is positive, that is,

$$4 \sin ^{2} x>1, \quad|\sin |>0.5$$

This is satisfied in the intervals $30^{\circ}<x<150^{\circ}$ and $210^{\circ}<x<330^{\circ}$. The excluded values $90^{\circ}$ and $270^{\circ}$ fall within these subintervals, so the solution is:

$$30^{\circ}<x<90^{\circ}, \quad 90^{\circ}<x<150^{\circ}, \quad 210^{\circ}<x<270^{\circ}, \quad 270^{\circ}<x<330^{\circ}$$

The last two intervals can be derived from the first two by adding $180^{\circ}$, so the general solution can be written more simply. In radians:

$$\frac{\pi}{6}+k \pi<x<\frac{\pi}{2}+k \pi \quad \text { and } \quad \frac{\pi}{2}+k \pi<x<\frac{5 \pi}{6}+k \pi$$

where $k$ is an integer.

János Kemenes (Budapest, Konyves Kalman g. III. o. t.)

Remarks. 1. $K$ can also be expressed in terms of the functions of $2 x$. By converting the difference and sum of sines into a product:

$$\begin{gathered} \sin ^{2} x-\sin ^{2} 3 x=(\sin x-\sin 3 x)(\sin x+\sin 3 x)= \\ =4 \cos 2 x \sin (-x) \sin 2 x \cos (-x)=-2 \sin ^{2} 2 x \cos 2 x, \text { and thus } \\ K=\sin ^{2} 2 x+\left(\sin ^{2} x-\sin ^{2} 3 x\right)=\sin ^{2} 2 x(1-2 \cos 2 x) \end{gathered}$$

This is positive if and only if

$$\begin{gathered} \sin ^{2} 2 x \neq 0 \text { and } \cos 2 x<0.5, \text { that is } \\ \frac{\pi}{3}+2 k \pi<2 x<\frac{5 \pi}{3}+2 k \pi \end{gathered}$$

The excluded value $2 x=n+2 k \pi$ falls within the obtained bounds, splitting the interval. Dividing by 2, we arrive at the above result.

Zoltán Demendy (Budapest, Hengersor uti g. IV. o. t.)

2. Several solved the problem graphically, using the usual representations of the terms of $K$. This method provides good orientation, but becomes uncertain near the critical points, i.e., where the two sides are equal. Without numerical examination, we cannot confidently state the magnitude relationships in the "small" neighborhood of these points.