Solve the equation $$\arcsin\bigg(\dfrac{x+1}{\sqrt{x^2+2x+2}}\bigg)-\arcsin\bigg(\dfrac{x}{\sqrt{x^2+1}}\bigg)=\dfrac{\pi}{4}$$
My solution: I converted this equation in terms of $\arctan$ and applied tangent to both sides, and I got my answer as $x=-1,0$.
But then one of my friends said that $x=2$ satisfies too the above equation, and the reason he gave is as follows:
$$\arcsin\bigg(\dfrac{3}{\sqrt{10}}\bigg)=\dfrac{\pi}{4}+\arcsin\bigg(\dfrac{2}{\sqrt{5}}\bigg)$$
and he applied sinus to both sides to obtain $\dfrac{3}{\sqrt{10}}=\dfrac{3}{\sqrt{10}}\:$.
Now I don't have any explanation for him. Can anyone here explain the reason behind this situation?
I plotted it in desmos, and I am getting $x=-1,0$ only.
Link to desmos
Best Answer
COMMENT.-Note that one of the two terms is equal to $0$ when $x=0$ and when $x=-1$. If the other term fits, then you do have solved directly the question. In fact this gives $$\arcsin\left(\frac{1}{\sqrt2}\right)-\arcsin(0)=\frac{\pi}{4}\\\arcsin(0)-\arcsin\left(-\frac{1}{\sqrt2}\right)=\frac{\pi}{4}$$ which is obviously true.