Solve: `|-2x^2+1+e^x+sinx|=|2x^2-1|+e^x+|sinx|,x in [0,2pi]`.

Explanation

`|-2x^(2)+1+e^(x)+sinx|=|2x^(2)-1|+e^(x)+|sinx|, x in [0,2pi]`
in the R.H.S., each term is positive and `e^(x) gt 0.` So,
`1-2x^(2) ge 0` and ` sin x ge 0`
or `x in [-(1)/(sqrt(2)),(1)/(sqrt(2))] and x in [0,pi]`
` :. x in [0,(1)/(sqrt(2))]`