treetopssan

Answered

2022-01-30

Find the roots (real or complex) of the equation $2{a}^{2}(1+\mathrm{cos}x)-{(1+{a}^{2}+{b}^{2})}^{2}=0,a,b\in \mathbb{R}$

ATTEMPT: Solving for x, we arrived with

$x=\mathrm{arccos}[\frac{{(1+{a}^{2}+{b}^{2})}^{2}}{2{a}^{2}}-1]$

However, I got stuck because of the required domain of$\mathrm{arccos}(\cdot )$ which is [-1,1] while the expression $\frac{{(1+{a}^{2}+{b}^{2})}^{2}}{2{a}^{2}}-1$ does not possibly be within this interval for arbitrary values of a and b. Sometimes I am tempted to say that no general solution exists even in C. Can anyone help me out?

ATTEMPT: Solving for x, we arrived with

However, I got stuck because of the required domain of

Answer & Explanation

Brynn Ortiz

Expert

2022-01-31Added 12 answers

What you should know is arccosx, if the domain is [-1,1], it has real value, and if the domain isnt,

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