Official solution
Given, \[f(x) = {{4{x^3} - 3{x^2}} \over 6} - 2\sin x + (2x - 1)\cos x\]
\[f'(x) = {{12{x^2} - 6x} \over 6} - 2\cos x + (2x - 1)( - \sin x) + \cos x(2)\]
\[ = (2{x^2} - x) - 2\cos x - 2x\sin x + \sin x + 2\cos x\]
\[ = 2{x^2} - x - 2x\sin x + \sin x\]
\[ = 2x(x - \sin x) - 1(x - \sin x)\]
\[f'(x) = (2x - 1)(x - \sin x)\]
for \[x > 0,x - \sin x > 0\]
\[x < 0,x - \sin x < 0\]
for \[x \in ( - \infty ,0] \cup \left[ {{1 \over 2},\infty } \right),f'(x) \ge 0\]
for \[x \in \left[ {0,{1 \over 2}} \right],f'(x) \le 0\]
Hence, f(x) increases in \[\left[ {{1 \over 2},\infty } \right)\].