Calculation: Let \[ f(x) = x^2 - x \sin x - \cos x \] Differentiating with respect to \( x \), we get \[ f'(x) = 2x - x \cos x - \sin x + \sin x \] Simplifying, \[ f'(x) = x(2 - \cos x) \] To find critical points, we set \( f'(x) = 0 \), \[ x(2 - \cos x) = 0 \] Since the maximum value of \( \cos x = 1 \), \[ (2 - \cos x) > 0 \] Thus, - \( f(x) \) is increasing when \( x > 0 \) - \( f(x) \) is decreasing when \( x < 0 \) Therefore, \[ \lim_{x \to \infty} f(x) = \infty, \quad \lim_{x \to -\infty} f(x) = -\infty, \quad \text{and} \quad f(0) = -1 \] It will cut x-axis at two points hence 2 solution