1. **State the problem:** We want to use trigonometric substitution to express the algebraic expression $$4 = \sqrt{25x^2 + 9}$$ as a function of $$\sec(\theta)$$, given that $$5x = 3\tan\theta$$ and $$0 \leq \theta \leq \frac{\pi}{2}$$. 2. **Recall the substitution and identities:** Given $$5x = 3\tan\theta$$, we can write $$x = \frac{3}{5}\tan\theta$$. The expression inside the square root is: $$25x^2 + 9 = 25\left(\frac{3}{5}\tan\theta\right)^2 + 9 = 25 \cdot \frac{9}{25} \tan^2\theta + 9 = 9\tan^2\theta + 9$$ 3. **Factor and simplify:** $$9\tan^2\theta + 9 = 9(\tan^2\theta + 1)$$ Recall the Pythagorean identity: $$1 + \tan^2\theta = \sec^2\theta$$ So, $$9(\tan^2\theta + 1) = 9\sec^2\theta$$ 4. **Take the square root:** $$\sqrt{25x^2 + 9} = \sqrt{9\sec^2\theta} = 3|\sec\theta|$$ Since $$0 \leq \theta \leq \frac{\pi}{2}$$, $$\sec\theta \geq 0$$, so: $$\sqrt{25x^2 + 9} = 3\sec\theta$$ 5. **Final expression:** The original expression $$4 = \sqrt{25x^2 + 9}$$ can be written as: $$4 = 3\sec\theta$$ **Answer:** $$\boxed{4 = 3\sec\theta}$$