Animated Solution for Mathematics - Trigonometry: The number of solutions of the equation $\sin x={\cos }^{2}x$ in the interval $(0,10)$ is ______.

Imagine you are standing on the shore, watching the rhythmic, oscillating waves of the ocean. In mathematics, the sine and cosine functions are the very embodiment of that rhythm.

Today, we are tackling a classic JEE Advanced problem: finding the number of solutions to $\sin x={\cos }^{2}x$ in the interval $(0,10)$.

At first glance, this equation feels like a tug-of-war between two different functions. We have $\sin x$ on one side and ${\cos }^{2}x$ on the other. This mixture is the primary obstacle.

To solve this, we must unify them. We know the fundamental Pythagorean identity: ${\cos }^{2}x=1-{\sin }^{2}x$.

By substituting this into our original equation, we transform the problem entirely. We are no longer dealing with two different functions; we are now looking at a single, unified expression in terms of $\sin x$.

Once we substitute the identity, our equation becomes $\sin x=1-{\sin }^{2}x$. Let us bring all the terms to one side to reveal the hidden structure.

Rearranging gives us ${\sin }^{2}x+\sin x-1=0$. Look closely at this; it is not just a trigonometric equation anymore, but a quadratic equation in disguise.

We can treat $\sin x$ as a single block, say $u$, turning the equation into $u^2+u-1=0$. Now, we can deploy our most reliable algebraic tool, the quadratic formula:

With $a=1$, $b=1$, and $c=-1$, we calculate the roots:

Filtering the Roots

In the heat of calculation, it is easy to forget the physical constraints of our functions. We have two potential values for $\sin x$: $\frac{-1+\sqrt{5}}{2}$ and $\frac{-1-\sqrt{5}}{2}$.

Remember, the sine function is strictly bounded between $-1$ and $1$. Let us evaluate our roots.

The value $\frac{-1-\sqrt{5}}{2}\approx -1.618$ is clearly outside our valid range, so we must reject it. The second value, $\frac{\sqrt{5}-1}{2}\approx 0.618$, falls perfectly within the range $[-1,1]$.

This is our target. We are now solving for $\sin x=\frac{\sqrt{5}-1}{2}$.

Now, let us step back and visualize the graph of $y=\sin x$. We are looking for the intersection of this sine wave with the horizontal line $y\approx 0.618$.

Our interval is $(0,10)$. To navigate this, we note that $\pi \approx 3.14$, $2\pi \approx 6.28$, and $3\pi \approx 9.42$. Since $4\pi \approx 12.56$, the value $x=10$ lies just after $3\pi$.

We draw our sine wave and mark the boundary at $x=10$. Now, we draw the horizontal line $y=0.618$.

In the first cycle, from $0$ to $2\pi$, the sine wave is positive in the interval $(0,\pi )$. The line $y=0.618$ intersects the curve twice in this region.

Moving to the second cycle, from $2\pi$ to $4\pi$, the sine wave is positive again in the interval $(2\pi ,3\pi )$. The line intersects the curve twice more.

Finally, in the interval $(3\pi ,10)$, the sine wave is negative, dipping below the x-axis. Our horizontal line $y=0.618$ is positive, so it will never touch the curve here.

Counting these intersections, we have two in the first cycle and two in the second. The total number of solutions is 4.