To solve the problem of finding all possible functions with the given derivative $f'(t) = \sin(3t) - \cos\left(\frac{t}{7}\right)$, we need to integrate each term with respect to $t$.

Step 1: Integrating $\sin(3t)$

The integral of $\sin(3t)$ is: $\int \sin(3t) \, dt = -\frac{1}{3} \cos(3t)$

Step 2: Integrating $-\cos\left(\frac{t}{7}\right)$

The integral of $\cos\left(\frac{t}{7}\right)$ is: $\int \cos\left(\frac{t}{7}\right) \, dt = 7 \sin\left(\frac{t}{7}\right)$ Thus, the integral of $-\cos\left(\frac{t}{7}\right)$ is: $-\int \cos\left(\frac{t}{7}\right) \, dt = -7 \sin\left(\frac{t}{7}\right)$

Step 3: Adding the results

By combining the integrals, the general form of $f(t)$ is: $f(t) = -\frac{1}{3} \cos(3t) - 7 \sin\left(\frac{t}{7}\right) + C$ where $C$ is the constant of integration.

Final Answer:

$f(t) = -\frac{1}{3} \cos(3t) - 7 \sin\left(\frac{t}{7}\right) + C$

Would you like further clarification on this solution?

Here are some related questions:

1. How do you compute integrals of trigonometric functions in general?
2. What role does the constant of integration play in indefinite integrals?
3. Can we use substitution to solve similar integrals?
4. What are the key steps for differentiating trigonometric functions?
5. How would the solution change if the derivative had more terms?

Tip: Always check your work by differentiating your result to verify it matches the given derivative.