How do I solve $|-2x^2+1+e^x+\sin x| = |2x^2-1|+e^x+|\sin x|$ where x belongs to [0,2π]?

Question

How do I solve $|-2x^2+1+e^x+\sin x| = |2x^2-1|+e^x+|\sin x|,$ where $x$ belongs to [0,2π]? My book solves it in this way: since RHS is positive, it concludes that $1- 2x^2 \ge 0$ and $ \sin x \ge 0$. Where have these results come from? How do I solve this problem?

Answer 1

We have $$|a+b+c|=|a|+|b|+|c|\quad\hbox{if and only if}\quad \hbox{$a,b,c$ all have the same sign}\ ,$$ where "same sign" is in the non-strict sense, that is, all ${}\ge0$ or all ${}\le0$. In this case $b=e^x$ is definitely non-negative, so $a=1-2x^2$ and $c=\sin x$ must be non-negative too.

To solve the problem, find all $x$ such that $1-2x^2\ge0$ and $\sin x\ge0$.