Let's carefully review the steps of the provided solution to the equation [tex]\( 3 \sin x + \cos^2 x = 2 \)[/tex]:
1. Start with the equation:
[tex]\[
3 \sin x + \cos^2 x = 2
\][/tex]
2. Apply the Pythagorean identity, [tex]\(\cos^2 x = 1 - \sin^2 x\)[/tex]:
[tex]\[
3 \sin x + (1 - \sin^2 x) = 2
\][/tex]
3. Simplify the equation:
[tex]\[
3 \sin x - \sin^2 x + 1 = 2
\][/tex]
4. Rearrange the equation:
[tex]\[
3 \sin x - \sin^2 x - 1 = 0
\][/tex]
5. Rewrite the quadratic equation in standard form:
[tex]\[
\sin^2 x - 3 \sin x + 1 = 0
\][/tex]
6. Factor the quadratic equation:
[tex]\[
(\sin x - 3)(\sin x + 1) = 0
\][/tex]
7. Solve the factored equation for [tex]\(\sin x\)[/tex]:
[tex]\[
\sin x - 3 = 0 \quad \text{or} \quad \sin x + 1 = 0
\][/tex]
[tex]\[
\sin x = 3 \quad \text{or} \quad \sin x = -1
\][/tex]
8. Find the inverse sine values:
[tex]\[
x = \sin^{-1}(3) \quad \text{or} \quad x = \sin^{-1}(-1)
\][/tex]
9. Simplify the solutions:
[tex]\[
x = \sin^{-1}(3) \quad \text{or} \quad x = \frac{3\pi}{2} + 2n\pi, \text{ since } \sin(\frac{3\pi}{2}) = -1
\][/tex]
Now let's analyze the solution:
- Step 2 correctly applies the Pythagorean identity [tex]\(\cos^2 x = 1 - \sin^2 x\)[/tex].
- Steps 3 through 5 correctly simplify and rearrange the equation.
- Step 6 appears to correctly factor the quadratic equation.
- Steps 7 and 8 correctly solve for [tex]\(\sin x\)[/tex].
The crucial point of error lies in Step 8. The sine function [tex]\(\sin x\)[/tex] has a range of [tex]\([-1, 1]\)[/tex]. Therefore, it is impossible for [tex]\(\sin x\)[/tex] to equal 3 because 3 is outside the range of the sine function.
Thus, the correct analysis shows that:
- The inverses were taken incorrectly because [tex]\(\sin x = 3\)[/tex] is not possible within the domain of sine function values.
Therefore, the best description of the solution is:
```
The inverses were taken incorrectly.
```
