求在 $0 \le x \le 7$ 范围内满足方程 $(2 + 3 \cos x)(4 - 5 \sin x) = 0$ 的所有 $x$ 的值。

1. Set each factor to zero to find the basic trigonometric equations

Since $(2 + 3 \cos x)(4 - 5 \sin x) = 0$, we have:
$2 + 3 \cos x = 0 \implies \cos x = -\frac{2}{3}$
or
$4 - 5 \sin x = 0 \implies \sin x = \frac{4}{5}$

2. Solve $\cos x = -\frac{2}{3}$ for $0 \le x \le 7$

Basic angle $\alpha = \cos^{-1}\left(\frac{2}{3}\right) \approx 0.84107$ rad.
Since $\cos x$ is negative, $x$ is in the 2nd and 3rd quadrants.
$x = \pi - 0.84107 \text{ or } \pi + 0.84107$
$x \approx 2.3005 \text{ or } 3.9827$
(The next solution $3\pi - 0.84107 \approx 8.58 > 7$)

3. Solve $\sin x = \frac{4}{5}$ for $0 \le x \le 7$

Basic angle $\beta = \sin^{-1}\left(\frac{4}{5}\right) \approx 0.92730$ rad.
Since $\sin x$ is positive, $x$ is in the 1st and 2nd quadrants.
$x = 0.92730 \text{ or } \pi - 0.92730$
$x \approx 0.92730 \text{ or } 2.2143$
(The next solution $2\pi + 0.92730 \approx 7.21 > 7$)

4. State the final values of $x$ rounded to 3 significant figures

$x = 0.927, 2.21, 2.30, 3.98$