The equation [tex]$3 \sin x + \cos^2 x = 2$[/tex] is solved below.
[tex]\[
\begin{array}{l}
3 \sin x + \cos^2 x = 2 \\
3 \sin x + \left(1 - \sin^2 x\right) = 2 \\
3 \sin x - \sin^2 x + 1 = 2 \\
3 \sin x - \sin^2 x - 1 = 0 \\
\sin^2 x - 3 \sin x + 1 = 0 \\
(\sin x - 3)(\sin x + 1) = 0 \\
\sin x - 3 = 0 \text{ or } \sin x + 1 = 0 \\
\sin x = 3 \text{ or } \sin x = -1 \\
x = \sin^{-1}(3) \text{ or } x = \sin^{-1}(-1) \\
x = \frac{2\pi}{3} + 2n\pi
\end{array}
\][/tex]

Which of the following best describes the solution?

A. The Pythagorean identity was applied incorrectly.
B. The equation was factored incorrectly.
C. The inverses were taken incorrectly.
D. There is no error in the solution.



Answer :

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.
```