Considering only the principal values of inverse functions, the set A=x0:^-1(2x)+^-1(3x)=4

Question

Considering only the principal values of inverse functions, the set $A=\left\{x\ge0:	an^{-1}(2x)+	an^{-1}(3x)=\dfrac{\pi}{4}ight\}$

Accepted Answer

$$\tan^{-1}(2x)+\tan^{-1}(3x)=\frac{\pi}{4}\ \Rightarrow\ \tan^{-1}\left(\frac{5x}{1-6x^2}\right)=\frac{\pi}{4}\ \Rightarrow\ \frac{5x}{1-6x^2}=1.$$

$$\Rightarrow 5x=1-6x^2\ \Rightarrow\ 6x^2+5x-1=0\ \Rightarrow\ (6x-1)(x+1)=0\ \Rightarrow\ x=\frac16\ (\text{since } x\ge0).$$

$A$ is a singleton.

Solutions · 1