Animated Solution for Mathematics - Differentiation: If $f(x)=\left|\begin{array}{ccc}\cos (2x)& \cos (2x)& \sin (2x)\\ \cos x& \cos x& \sin x\\ -\sin x& \sin x& \cos x\end{array}\right|$, then

Welcome, fellow traveler on the path of JEE mastery. Today, we face a problem that looks like a fortress of complexity: a $3\times 3$ determinant filled with trigonometric functions.

It is easy to feel overwhelmed, but remember: every complex structure is built from simple, elegant bricks. Our goal is to analyze the function $f(x)$ defined as:

To dismantle this fortress, we perform what I call 'algebraic surgery.' We expand the determinant along the first row ($R_1$). This is the standard procedure, and it is our most reliable tool.

We take the first element, $\cos (2x)$, and multiply it by its corresponding $2\times 2$ minor. Then, we subtract the second element, $\cos (2x)$, multiplied by its minor. Finally, we add the third element, $\sin (2x)$, multiplied by its minor.

The expression becomes:

Now, we evaluate those $2\times 2$ minors. This is where the magic happens.

For the first minor, we calculate $(\cos x\cdot \cos x)-(\sin x\cdot \sin x)={\cos }^{2}x-{\sin }^{2}x$. For the second, we get $(\cos x\cdot \cos x)-(\sin x\cdot -\sin x)={\cos }^{2}x+{\sin }^{2}x$. For the third, we get $(\cos x\cdot \sin x)-(\cos x\cdot -\sin x)=2\sin x\cos x$.

Suddenly, the fog clears. We recall our fundamental identities: ${\cos }^{2}x-{\sin }^{2}x=\cos (2x)$, ${\cos }^{2}x+{\sin }^{2}x=1$, and $2\sin x\cos x=\sin (2x)$.

Substituting these back, the function collapses into something beautiful:

We have reduced a terrifying determinant to $f(x)=1-\cos (2x)$. Now, let us analyze its motion.

To find the critical points, we differentiate:

Setting $f^{\prime }(x)=0$ gives us $\sin (2x)=0$, which implies $2x=n\pi$, or $x=\frac{n\pi }{2}$. Within the interval $(-\pi ,\pi )$, the valid values for $n$ are $-1,0,1$, giving us exactly three critical points: $x=-\frac{\pi }{2},0,\frac{\pi }{2}$.

Finally, we check the extrema. Since the range of $\cos (2x)$ is $[-1,1]$, the minimum of $f(x)=1-\cos (2x)$ occurs when $\cos (2x)$ is at its maximum of $1$.

Thus, $f_{min}=1-1=0$, which occurs at $x=0$.

We have conquered the problem! We found that $f^{\prime }(x)=0$ at exactly three points and that the minimum is at $x=0$. Keep this confidence with you; you are capable of solving any problem that comes your way.