How do you solve \[\arcsin \left( x \right) + \arctan \left( x \right) = 0\] ?

Hint: Given is an equation of two trigonometric functions in inverse form. We need to solve the same problem, which is to find the value of x. for this we need to know the domain of the functions as well. We will convert the tan function in the sin function and then we will proceed. After that we will find the value of x depending on the domain of sin function. Let’s solve it!

Complete step by step solution:
Given that,
\[\arcsin \left( x \right) + \arctan \left( x \right) = 0\]
Now we can write this as,
\[\arcsin \left( x \right) = - \arctan \left( x \right)\]
Removing the common arc function,
\[sin\left( x \right) = - tan\left( x \right)\]
Now we know that \[\tan x = \dfrac{{\sin x}}{{\cos x}}\]
So we can write,
\[sin\left( x \right) = - \dfrac{{\sin x}}{{\cos x}}\]
Taking it on left side,
\[sin\left( x \right) + \dfrac{{\sin x}}{{\cos x}} = 0\]
Taking the LCM we get,
\[\dfrac{{\sin x.\cos x + \sin x}}{{\cos x}} = 0\]
Now transposing denominator we get,
\[\sin x.\cos x + \sin x = 0\]
Now taking sinx common we can write,
\[\sin x\left( {\cos x + 1} \right) = 0\]
Now equating the terms separately to zero we get,
\[\sin x = 0or\left( {\cos x + 1} \right) = 0\]
For second bracket we can write,
\[\sin x = 0or\cos x = - 1\]
So we can conclude,
\[x = 0\] since sin zero degrees is 0.
\[x = \pi \] since \[\cos \pi = - 1\]
So \[x = 0,\pi \]
But since the domain of arcsin is between -1 and 1. So we can just take zero as the correct solution that is \[x = 0\].

Note: Note that in these types of problems we first need to simplify the functions. And then we will start finding the angle. Also note that arc is nothing but an inverse function. We chose the sin function domain because in our problem we have the same function.