0\\
x(3x-2)>0\\
x\in(-\infty;0) \ \cup \ (\frac{2}{3};\infty)\\\\
log_{\frac{1}{2}} (3x^2-2x) \geq 0\\
log_{2}(3x^2-2x) \leq 0\\
3x^2-2x \leq 1\\
x\in[-\frac{1}{3};1]\\\\
\\\\
x\in[-\frac{1}{3};0) \ \cup (\frac{2}{3};1]
" alt="3x^2-2x>0\\
x(3x-2)>0\\
x\in(-\infty;0) \ \cup \ (\frac{2}{3};\infty)\\\\
log_{\frac{1}{2}} (3x^2-2x) \geq 0\\
log_{2}(3x^2-2x) \leq 0\\
3x^2-2x \leq 1\\
x\in[-\frac{1}{3};1]\\\\
\\\\
x\in[-\frac{1}{3};0) \ \cup (\frac{2}{3};1]
" align="absmiddle" class="latex-formula">