1. **State the problem:** Find the equation of the tangent line to the function $F(x) = -\cos(3x) + 4$ at the point where $x = \frac{\pi}{2}$. 2. **Recall the formula for the tangent line:** $$y - y_1 = m(x - x_1)$$ where $m$ is the slope of the tangent line at $x_1$, and $(x_1, y_1)$ is the point of tangency. 3. **Find $y_1$ by evaluating $F\left(\frac{\pi}{2}\right)$:** $$y_1 = -\cos\left(3 \times \frac{\pi}{2}\right) + 4 = -\cos\left(\frac{3\pi}{2}\right) + 4$$ Since $\cos\left(\frac{3\pi}{2}\right) = 0$, we get $$y_1 = -0 + 4 = 4$$ 4. **Find the slope $m$ by differentiating $F(x)$:** $$F'(x) = \frac{d}{dx} \left(-\cos(3x) + 4\right) = -(-\sin(3x)) \times 3 = 3\sin(3x)$$ 5. **Evaluate the slope at $x = \frac{\pi}{2}$:** $$m = F'\left(\frac{\pi}{2}\right) = 3 \sin\left(3 \times \frac{\pi}{2}\right) = 3 \sin\left(\frac{3\pi}{2}\right)$$ Since $\sin\left(\frac{3\pi}{2}\right) = -1$, we get $$m = 3 \times (-1) = -3$$ 6. **Write the tangent line equation using the point-slope form:** $$y - 4 = -3 \left(x - \frac{\pi}{2}\right)$$ 7. **Simplify the equation:** $$y - 4 = -3x + \frac{3\pi}{2}$$ $$y = -3x + \frac{3\pi}{2} + 4$$ **Final answer:** $$\boxed{y = -3x + \frac{3\pi}{2} + 4}$$