Matematické Fórum

stuart clark · 17. 03. 2020 10:03

If the equation $\csc^2(x)+\csc(x)-k-1=0$ has at least one real solution. Then range of $k$ is

auditor · 17. 03. 2020 19:51

↑ stuart clark:

The equation has at least one real solution for k equal or higher than -1.

stuart clark · 18. 03. 2020 10:25

Thanks ↑ auditor: got it.

Let $\csc x=a.$ Then equation is $f(a)=a^2+a-1=k$ has at least one solution in $|a|\geq 1.$

Here $f(a)=a^2+a-1$ is upward parabola and $f(-1)=-1$ and $f(1)=1$ and $\lim_{a\rightarrow \pm \infty}f(a)\rightarrow \infty$

So $f(a)=k\geq -1$ for at least one real $|a|\geq 1.$

Matematické Fórum

#1 17. 03. 2020 10:03

Range of k

#2 17. 03. 2020 19:51

Re: Range of k

#3 18. 03. 2020 10:25

Re: Range of k

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