Step-by-step Solution:

We are given two equations: \[\] 1. \( e^x \sin x = 1 \) \[\] 2. \( e^x \cos x = -1 \) \[\] We are tasked with determining how many real roots the equation \( e^x \cos x = -1 \) has between any two real roots of the equation \( e^x \sin x = 1 \). \[\] Step 1: Analyze the first equation \( e^x \sin x = 1 \) \[\] The equation \( e^x \sin x = 1 \) describes a curve where the product of the exponential function \( e^x \) and the sine function \( \sin x \) is equal to 1. Let's examine the behavior of this equation: \[\] As \( x \to \infty \), \( e^x \) grows exponentially, and \( \sin x \) oscillates between \(-1\) and \(1\). Therefore, \( e^x \sin x \) will continue to grow and oscillate, and will achieve a value of 1 at certain points. \[\] As \( x \to -\infty \), \( e^x \) tends towards zero, so \( e^x \sin x \) will be very small and cannot be equal to 1. \[\] The equation \( e^x \sin x = 1 \) will therefore have infinitely many real solutions for \( x \) because the oscillations of \( \sin x \) combined with the exponential growth of \( e^x \) allow \( e^x \sin x \) to reach 1 at multiple points. \[\] Step 2: Analyze the second equation \( e^x \cos x = -1 \) \[\] The equation \( e^x \cos x = -1 \) describes a curve where the product of the exponential function \( e^x \) and the cosine function \( \cos x \) is equal to \(-1\). Let's analyze this equation: \[\] As \( x \to \infty \), \( e^x \) grows exponentially, and \( \cos x \) oscillates between \(-1\) and \(1\). Thus, \( e^x \cos x \) will grow without bound and oscillate. \[\] As \( x \to -\infty \), \( e^x \) approaches zero, and \( e^x \cos x \) will approach zero and cannot equal \(-1\). \[\] So, \( e^x \cos x = -1 \) also has infinitely many real solutions as \( x \) increases. \[\] Step 3: Count the solutions between two real roots of \( e^x \sin x = 1 \) \[\] Since \( e^x \sin x = 1 \) has infinitely many real solutions, it follows that there are infinitely many intervals between two consecutive real roots of \( e^x \sin x = 1 \). \[\] Now, consider the behavior of the equation \( e^x \cos x = -1 \) on such an interval: \[\] The function \( e^x \cos x \) oscillates with the exponential growth of \( e^x \) and the periodic oscillations of \( \cos x \). \[\] Between any two real roots of \( e^x \sin x = 1 \), the function \( e^x \cos x \) will undergo at least one complete oscillation, meaning the equation \( e^x \cos x = -1 \) will have at least one solution in each such interval. \[\] Conclusion: \[\] Between any two real roots of the equation \( e^x \sin x = 1 \), the equation \( e^x \cos x = -1 \) have at least one solution.