Animated Solution for Mathematics - Matrices and Determinants: The solutions of the equation $\left|\begin{array}{ccc}1+{\sin }^{2}x& {\sin }^{2}x& {\sin }^{2}x\\ {\cos }^{2}x& 1+{\cos }^{2}x& {\cos }^{2}x\\ 4\sin 2x& 4\sin 2x& 1+4\sin 2x\end{array}\right|=0,(0<x<\pi )$ are:

Imagine you are standing before a wall of numbers, a complex determinant that seems designed to intimidate. It is easy to feel overwhelmed by the ${\sin }^{2}x$ and ${\cos }^{2}x$ terms, but remember, in the world of JEE Advanced, every complex structure hides a simple, elegant core.

We are given the equation:

The constraint provided is $0<x<\pi$. The key to unlocking this puzzle lies in recognizing the symmetry.

By applying the row operation $R_1\to R_1+R_2+R_3$, we invoke the fundamental identity ${\sin }^{2}x+{\cos }^{2}x=1$. Suddenly, the first row transforms into $2+4\sin 2x$ across all three columns.

With the first row now consisting of identical terms, we can factor out $(2+4\sin 2x)$ completely. This leaves us with:

To simplify further, we use column operations $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$. This turns our matrix into a lower triangular form, where the determinant is simply the product of the diagonal elements.

Since the diagonal elements are now $1$, the entire determinant simplifies to just $1$. We are left with the elegant equation:

Now, we move to the final phase. We isolate the sine term: $4\sin 2x=-2$, which simplifies to:

The original constraint is $0<x<\pi$, which means our angle $2x$ must lie in the range $0<2x<2\pi$. This is a full rotation on the unit circle.

We are looking for where the sine value is $-\frac{1}{2}$. This occurs in the third and fourth quadrants.

The first solution is $2x=\pi +\frac{\pi }{6}=\frac{7\pi }{6}$, and the second is $2x=2\pi -\frac{\pi }{6}=\frac{11\pi }{6}$.

Finally, we solve for $x$ by dividing by $2$. The resulting values are:

Both values fall perfectly within our range. You have successfully navigated the complexity and found the truth hidden within the numbers.