44) \( -3 \sin x \cos x<0 \)

To solve the inequality \( -3 \sin x \cos x < 0 \), we first simplify it to \( \sin x \cos x > 0 \) (since multiplying by -1 reverses the inequality). Knowing that \( \sin x \cos x = \frac{1}{2} \sin(2x) \), we deduce that our inequality reduces to \( \sin(2x) > 0 \). This means \( 2x \) must lie in intervals where the sine function is positive, which occurs in the ranges \( (0, \pi) \) and \( (2\pi, 3\pi) \). Thus, for the original \( x \), the solution converts to: - \( 0 < 2x < \pi \) leading to \( 0 < x < \frac{\pi}{2} \) - \( 2\pi < 2x < 3\pi \) giving \( \pi < x < \frac{3\pi}{2} \) So, \( x \) values satisfying \( -3 \sin x \cos x < 0 \) are in \( (0, \frac{\pi}{2}) \) and \( (\pi, \frac{3\pi}{2}) \).