To find the number of solutions for the equation ${x^2} - x\sin x - \cos x = 0$, we can analyze the behavior of the function $f(x) = {x^2} - x\sin x - \cos x$ using calculus.

Step 1: Define the function

Let the given equation be represented by a function $f(x)$: $f(x) = {x^2} - x\sin x - \cos x$ The number of solutions to the equation is the number of roots of $f(x)=0$.

Step 2: Find the first derivative of the function

We differentiate $f(x)$ with respect to $x$ to find its critical points and determine its intervals of increase and decrease. $f'(x) = \frac{d}{{dx}}({x^2} - x\sin x - \cos x)$ Using the product rule for the term $-x\sin x$, we get: $f'(x) = 2x - (1 \cdot \sin x + x \cdot \cos x) - ( - \sin x)$ $f'(x) = 2x - \sin x - x\cos x + \sin x$ $f'(x) = 2x - x\cos x$ Factoring out $x$, we have: $f'(x) = x(2 - \cos x)$

Step 3: Analyze the sign of the first derivative

The range of the cosine function is $-1 \le \cos x \le 1$. Therefore, the term $(2 - \cos x)$ will always be positive, as its value ranges from $2-1=1$ to $2-(-1)=3$. $1 \le (2 - \cos x) \le 3$ Since $(2 - \cos x) > 0$ for all $x \in \mathbb{R}$, the sign of $f'(x)$ is determined solely by the sign of $x$.

• For $x > 0$, $f'(x) > 0$, which means $f(x)$ is strictly increasing.
• For $x < 0$, $f'(x) < 0$, which means $f(x)$ is strictly decreasing.
• For $x = 0$, $f'(x) = 0$, which indicates a critical point at $x=0$.

Step 4: Determine the nature of the critical point

Since the function is decreasing for $x < 0$ and increasing for $x > 0$, the critical point at $x=0$ must be a global minimum.

Step 5: Evaluate the function at the minimum and at the boundaries

Let's find the value of the function at its global minimum, $x=0$: $f(0) = {0^2} - 0 \cdot \sin(0) - \cos(0) = 0 - 0 - 1 = -1$ So, the minimum value of the function is -1.

Now, let's examine the behavior of $f(x)$ as $x$ approaches $\pm\infty$: As $x \to \infty$, the $x^2$ term dominates the other terms ($|x\sin x| \le |x|$ and $|\cos x| \le 1$). $\lim_{{x \to \infty}} f(x) = \lim_{{x \to \infty}} ({x^2} - x\sin x - \cos x) = \infty$ Similarly, as $x \to -\infty$: $\lim_{{x \to -\infty}} f(x) = \lim_{{x \to -\infty}} ({x^2} - x\sin x - \cos x) = \infty$

Step 6: Conclude the number of roots

We can summarize the behavior of the continuous function $f(x)$:

1. In the interval $(-\infty, 0)$, the function $f(x)$ is strictly decreasing from $\infty$ to its minimum value of $f(0) = -1$. Since the function goes from a positive value to a negative value, by the Intermediate Value Theorem, it must cross the x-axis exactly once. So, there is one root in $(-\infty, 0)$.
2. In the interval $(0, \infty)$, the function $f(x)$ is strictly increasing from its minimum value of $f(0) = -1$ to $\infty$. Since the function goes from a negative value to positive values, it must cross the x-axis exactly once. So, there is one root in $(0, \infty)$.

At $x=0$, $f(0)=-1 \ne 0$, so $x=0$ is not a root.

Therefore, the total number of points for which $f(x)=0$ is $1 + 1 = 2$.

The graph of the function starts at $+\infty$ for $x \to -\infty$, decreases to a minimum of $-1$ at $x=0$, and then increases back to $+\infty$ for $x \to \infty$. This confirms it must intersect the x-axis at exactly two points.