Number of Intersection Points of Exponential and Trigonometric Curves

To find the number of points of intersection of the curves $C_1$ and $C_2$ for $x \in [0, 10\pi]$, we set their $y$-values equal to each other:

$e^{-x} = e^{-x}(\sin x + \cos x)$

Since $e^{-x} > 0$ for all real $x$, we can divide both sides by $e^{-x}$:

$\sin x + \cos x = 1$

To solve this trigonometric equation, we rewrite the left-hand side in the harmonic form by multiplying and dividing by $\sqrt{2}$:

$\sqrt{2} \left(\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x\right) = 1$

$\sqrt{2} \sin\left(x + \frac{\pi}{4}\right) = 1$

$\sin\left(x + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

The general solutions for this equation are given by:

$x + \frac{\pi}{4} = 2k\pi + \frac{\pi}{4} \quad \text{or} \quad x + \frac{\pi}{4} = 2k\pi + \frac{3\pi}{4}, \quad \text{where } k \in \mathbb{Z}$

Simplifying these two sets of solutions gives:

1. $x = 2k\pi$
2. $x = 2k\pi + \frac{\pi}{2}$

Now, we find the values of $x$ that lie within the given interval $[0, 10\pi]$:

• From $x = 2k\pi$: For $k = 0, 1, 2, 3, 4, 5$, the values are: $x \in \{0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi\}$ (Total of 6 solutions)

• From $x = 2k\pi + \frac{\pi}{2}$: For $k = 0, 1, 2, 3, 4$, the values are: $x \in \left\{\frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, \frac{13\pi}{2}, \frac{17\pi}{2}\right\}$ (Total of 5 solutions)

Combining both sets of solutions in increasing order, we get:

$\alpha_1 = 0, \quad \alpha_2 = \frac{\pi}{2}, \quad \alpha_3 = 2\pi, \quad \alpha_4 = \frac{5\pi}{2}, \quad \alpha_5 = 4\pi, \quad \alpha_6 = \frac{9\pi}{2},$ $\alpha_7 = 6\pi, \quad \alpha_8 = \frac{13\pi}{2}, \quad \alpha_9 = 8\pi, \quad \alpha_{10} = \frac{17\pi}{2}, \quad \alpha_{11} = 10\pi$

Thus, the total number of intersection points $n$ is $6 + 5 = 11$.