Animated Solution for Mathematics - Trigonometry: The number of solutions of the equation $2\theta -{\cos }^2\theta +\sqrt{2}=0$ in $\mathbb{R}$ is equal to ______.

The equation provided is $2\theta -{\cos }^2\theta +\sqrt{2}=0$. At first glance, this appears to be a standard algebraic challenge, but the presence of both linear and trigonometric terms identifies it as a transcendental equation.

In the context of JEE Advanced, such problems test your ability to visualize function behavior rather than mere symbolic manipulation. We must look for the "soul" of these functions to determine the nature of the solution.

To simplify our perspective, we rearrange the equation into a form that highlights the relationship between two distinct functions:

We define the left side as $f(\theta )={\cos }^2\theta$ and the right side as $g(\theta )=2\theta +\sqrt{2}$. The solutions to the original equation are the points where these two graphs intersect.

First, consider $f(\theta )={\cos }^2\theta$. For any real $\theta$, the value of $\cos \theta$ is trapped in the interval $[-1,1]$. Consequently, $f(\theta )$ is strictly bounded such that $0\le f(\theta )\le 1$.

Next, consider $g(\theta )=2\theta +\sqrt{2}$. This is a straight line with a slope of $2$ and a $y$-intercept of $\sqrt{2}\approx 1.414$.

For any $\theta \ge 0$, we observe that $g(\theta )\ge \sqrt{2}\approx 1.414$. Since the wave $f(\theta )$ can never exceed $1$, the line $g(\theta )$ remains strictly above the wave for all positive $\theta$. Thus, no intersection exists for $\theta \ge 0$.

Since no solutions exist for $\theta \ge 0$, we investigate the negative domain. At $\theta =0$, we have $g(0)=\sqrt{2}>f(0)=1$.

Let us test $\theta =-\frac{\pi }{2}$:

At this point, the line is at $-1.73$, which is below the wave at $0$. Because both functions are continuous, the Intermediate Value Theorem guarantees that the line must cross the wave at least once in the interval $(-\frac{\pi }{2},0)$.

To determine if the solution is unique, we examine the derivatives. The slope of the line is $g^{\prime }(\theta )=2$. The slope of the wave is:

The maximum value of $|f^{\prime }(\theta )|$ is $1$. This indicates that the wave is never steeper than $1$, while the line maintains a constant, steeper slope of $2$.

Because the line is always steeper than the wave, once it crosses the wave, it will diverge from it indefinitely. It is impossible for the wave to catch up to the line again.

Conclusion: The intersection is unique. The total number of solutions is 1.