Identifying the Problem
Given equation: 3sin(theta) + 4cos(theta) = A*sin(theta + phi)

Finding A
We need to find the value of A in the given equation. To do this, we can use trigonometric identities to rewrite the equation in terms of sin(theta + phi).
Using the identity sin(A + B) = sinAcosB + cosAsinB, we can rewrite sin(theta + phi) as sin(theta)cos(phi) + cos(theta)sin(phi).

Substituting Values
Now, substitute sin(theta + phi) with the rewritten expression in the given equation:
3sin(theta) + 4cos(theta) = A(sin(theta)cos(phi) + cos(theta)sin(phi))
Expand the right side of the equation:
3sin(theta) + 4cos(theta) = A*sin(theta)cos(phi) + A*cos(theta)sin(phi)

Comparing Coefficients
Now, compare the coefficients of sin(theta) and cos(theta) on both sides of the equation:
3 = A*cos(phi) (coefficient of sin(theta))
4 = A*sin(phi) (coefficient of cos(theta))

Solving for A
To find A, we can square both equations and add them together:
(3)^2 + (4)^2 = A^2(cos(phi))^2 + A^2(sin(phi))^2
Simplify the equation:
25 = A^2
Therefore, A = 5.

Conclusion
The value of A in the given equation is 5.