How to solve $|-2x^2+1+e^x+\sin(x)|=|2x^2-1|+e^x+|\sin(x)|$?

Question

How to solve $|-2x^2+1+e^x+\sin(x)|=|2x^2-1|+e^x+|\sin(x)|$ ?

I've solved equations like $|a|+|b|=|a+b|$ where the condition must be that $a$, $b$ must be of same sign. But in case of three terms like the one above what should be the condition?

Answer 1

If $x$ is a real number, then all three quantities must be positive and intersection of all three sets will be final value of $x$ i.e $2x^2-1>0$ & $\sin(x)>0$. So $x\in[\frac{1}{\sqrt{2}},\frac{\pi}{2}] \cup [2n\pi,2n\pi+\frac{\pi}{2}]$ and same for the negative values of $x$.

Hope this is helpful

Answer 2

You didn't specify what $x$ is. The following only works if $x$ is a real number.

Hint: $|e^x| = e^x$ and $|2x^2-1| = |1-2x^2|$. So your equation is of the form $|a+b+c| = |a|+|b|+|c|$.

Answer 3

Hint:

You can first detect the values where the arguments of the absolute values change sign and rewrite the equation in the corresponding intervals (using $|x|=x$ or $|x|=-x$ where appropriate).

In your problem, the interesting bounds are $\pm\frac1{\sqrt2}$ and $k\pi$. The roots of the LHS are much harder to get.

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From a plot of the two sides, you can see that the equation is an identity in some interval and has no solutions elsewhere.

Answer 4

To avoid calculations and achieve at a direct answer, what you could do is compare the domains of the expressions on the left hand side and right hand side of the equations and then take out the common values of x from both the domains. That would be your answer.