Animated Solution for Mathematics - Inverse Trigonometric Functions: Domain of definition of the function $f(x)=\sqrt{{\sin }^{-1}(2x)+\pi /6}$ for real valued $x$, is

Welcome, future engineers! Today, we are going to dissect a problem that looks intimidating but is actually a beautiful exercise in logical layering. We are looking for the domain of the function $f(x)=\sqrt{{\sin }^{-1}(2x)+\frac{\pi }{6}}$.

When you encounter a composite function like this, do not panic. Think of it as a Russian Matryoshka doll or an onion. To find the domain, we must peel back the layers and ensure that every single part of the function is 'happy'—meaning it is mathematically defined.

The Square Root Constraint

The outermost layer is the square root. In the realm of real numbers, a square root is a picky eater; it refuses to accept negative values. Therefore, the expression inside must be non-negative.

We write this as:

This is our first law. To solve for $x$, we isolate the inverse sine term by shifting the constant to the other side:

Now, we need to strip away the ${\sin }^{-1}$ operator. We apply the sine function to both sides. Here is the critical moment where many students stumble: does the inequality flip?

Because $\sin (\theta )$ is a strictly increasing function in its principal domain, the inequality sign remains steadfast. We get:

Recalling our trigonometric identities, we know that $\sin (-\theta )=-\sin (\theta )$, and $\sin (\frac{\pi }{6})=\frac{1}{2}$. Thus, $\sin (-\frac{\pi }{6})=-\frac{1}{2}$.

Our inequality becomes $2x\ge -\frac{1}{2}$, which simplifies to $x\ge -\frac{1}{4}$. This is our first boundary.

The Hidden Constraint of the Inner Function

But wait! We are not done. We must respect the inner function, ${\sin }^{-1}(2x)$.

The inverse sine function is not defined for just any input; it demands that its argument lies strictly within the closed interval $[-1,1]$. This is the 'hidden' constraint that catches many students off guard.

We must satisfy:

Dividing this entire compound inequality by $2$, we obtain the second constraint:

Now, we have two conditions that must be satisfied simultaneously. We have the 'Square Root' condition ($x\ge -\frac{1}{4}$) and the 'Inverse Sine' condition ($-\frac{1}{2}\le x\le \frac{1}{2}$).

To find the true domain, we must find the intersection of these two sets on the number line. Imagine a number line where the first condition highlights everything from $-1/4$ to infinity, and the second condition highlights the segment from $-1/2$ to $1/2$.

The overlap, the region where both conditions are true, is the interval $[-\frac{1}{4},\frac{1}{2}]$.

And there you have it! By systematically peeling the layers and respecting the domain of each component, we have arrived at the final domain: $[-\frac{1}{4},\frac{1}{2}]$.

Mathematics is not about memorizing steps; it is about understanding the constraints of the universe we are working in. Keep practicing, keep questioning, and keep falling in love with the logic!