69. (grades 10-11) Solve the inequality $\sin x \sin 2 x<\sin 3 x \sin 4 x$, if $0<x<\frac{\pi}{2}$.

## 1965 year
69. Replacing the product of sines with the differences of the corresponding cosines, we get $\cos 3 x > \cos 7 x$ or $2 \sin 2 x \sin 5 x > 0$. Since $0 < x < \pi$, $\sin 2x > 0$. Therefore, the inequality $2 \sin 2 x \sin 5 x > 0$ holds under the condition $0 < x < \pi$ if $\sin 5x > 0$. Since $\sin 5x > 0$ for $0 < x < \pi$, negative values are not considered.

## 1965 year