Решите уравнение arccos(2x^2-1)=arccoc(3x+1)

ОДЗ:
$\left \{ {{-1 \leq 2x^2-1 \leq 1} \atop {-1 \leq 3x+1 \leq 1}} \right. \ \ \textless \ =\ \textgreater \ \ \left \{{{0 \leq 2x^2 \leq 2 } \atop {-2 \leq 3x \leq 0}} \right. \ \ \textless \ =\ \textgreater \ \ \left \{ {{0 \leq x^2 \leq 1} \atop {- \frac{2}{3} \leq x \leq 0 }} \right. \ \ \textless \ =\ \textgreater \ \left \{ {{ x^2 \leq 1} \atop {- \frac{2}{3} \leq x \leq 0 }} \right. \\ \\ \left \{ {{-1 \leq x \leq 1} \atop {- \frac{2}{3} \leq x \leq 0 }} \right. \ \ \textless \ =\ \textgreater \ \ -\frac{2}{3} \leq x \leq 0$

$arccos(2x^2-1)=arccos(3x+1) \\ \\ 2x^2-1=3x+1 \\ \\ 2x^2-3x-2=0 \\ \\ D=9+4*2*2=9+16=25=5^2 \\ \\ x_1= \frac{3-5}{2*2}= -\frac{1}{2} \\ \\$
$x_2= \frac{3+5}{2*2} =2$ - не удовлетворяет ОДЗ.

Ответ: -1/2