To find the equation of the tangent, we first need to find the derivative of \(y = x \sin 2x\) using the product rule.

The product rule states that if \(y = u \cdot v\), then \(\frac{dy}{dx} = u'v + uv'\).

Let \(u = x\) and \(v = \sin 2x\).

Then \(u' = 1\) and \(v' = 2 \cos 2x\) (using the chain rule).

So, \(\frac{dy}{dx} = 1 \cdot \sin 2x + x \cdot 2 \cos 2x = \sin 2x + 2x \cos 2x\).

Now, evaluate the derivative at \(x = \frac{1}{4} \pi\):

\(\frac{dy}{dx} \bigg|_{x = \frac{1}{4} \pi} = \sin \left( \frac{1}{2} \pi \right) + 2 \cdot \frac{1}{4} \pi \cdot \cos \left( \frac{1}{2} \pi \right)\).

Since \(\sin \left( \frac{1}{2} \pi \right) = 1\) and \(\cos \left( \frac{1}{2} \pi \right) = 0\), we have:

\(\frac{dy}{dx} \bigg|_{x = \frac{1}{4} \pi} = 1 + 0 = 1\).

The slope of the tangent is 1.

Now, find the point on the curve at \(x = \frac{1}{4} \pi\):

\(y = \frac{1}{4} \pi \sin \left( \frac{1}{2} \pi \right) = \frac{1}{4} \pi \cdot 1 = \frac{1}{4} \pi\).

The point is \(\left( \frac{1}{4} \pi, \frac{1}{4} \pi \right)\).

The equation of the tangent line is:

\(y - \frac{1}{4} \pi = 1 \left( x - \frac{1}{4} \pi \right)\).

Simplifying gives:

\(y = x\).