The equation given is:

$\sin^2(x) + 3\cos(x) - 1 = 2$

Step-by-step solution:

1. Rearrange the equation:

   Subtract 2 from both sides to simplify the equation:

   $\sin^2(x) + 3\cos(x) - 1 - 2 = 0$

   This simplifies to:

   $\sin^2(x) + 3\cos(x) - 3 = 0$

2. Use the Pythagorean identity:

   We know that $\sin^2(x) = 1 - \cos^2(x)$. Substitute this into the equation:

   $(1 - \cos^2(x)) + 3\cos(x) - 3 = 0$

   Simplify the equation:

   $1 - \cos^2(x) + 3\cos(x) - 3 = 0$

   $-\cos^2(x) + 3\cos(x) - 2 = 0$

   Multiply the equation by -1 to make the leading coefficient of $\cos^2(x)$ positive:

   $\cos^2(x) - 3\cos(x) + 2 = 0$

3. Solve the quadratic equation:

   Now solve for $\cos(x)$ using the quadratic formula. This is a standard quadratic equation in the form:

   $\cos^2(x) - 3\cos(x) + 2 = 0$

   Factor it:

   $(\cos(x) - 1)(\cos(x) - 2) = 0$

   So, the solutions are:

   $\cos(x) = 1 \quad \text{or} \quad \cos(x) = 2$

4. Consider the valid solutions:

   • $\cos(x) = 1$ is valid because the cosine function can take the value 1. This happens when $x = 0 + 2k\pi$, where $k$ is any integer.
   • $\cos(x) = 2$ is not valid because the cosine function can only take values between -1 and 1.

5. Final solution:

   The exact solution is $x = 2k\pi$ for any integer $k$.

Answer:

The correct choice is $2k\pi$.

Would you like more details or have any questions? Here are five related questions to further your understanding:

1. How would you solve this equation if the quadratic had no real roots?
2. What is the general form of the quadratic formula, and how does it apply here?
3. Why is $\cos(x) = 2$ not a valid solution in trigonometry?
4. How does the periodic nature of cosine affect the general solution for $x$?
5. Could there be additional solutions if we considered negative angles?

Tip: Always check the domain of trigonometric functions to ensure that all solutions are valid in the context of the equation!