Let's review the steps in the solution provided to see if there are any errors:
-
The original equation is \( 3 \sin x + \cos^2 x = 2 \).
-
The substitution of \(\cos^2 x\) using the identity \(\cos^2 x = 1 - \sin^2 x\) yields: \[ 3 \sin x + (1 - \sin^2 x) = 2 \]
-
Rearranging gives: \[ 3 \sin x - \sin^2 x + 1 = 2 \] Simplifying this leads to: \[ 3 \sin x - \sin^2 x - 1 = 0 \] This becomes: \[ \sin^2 x - 3 \sin x + 1 = 0 \]
-
The quadratic equation \(\sin^2 x - 3 \sin x + 1 = 0\) can be factored (it actually cannot be factored nicely): \[ (\sin x - 3)(\sin x - 1) = 0 \]
However, \(\sin x - 3 = 0\) leads to an invalid solution because the sine function can only take values between -1 and 1.
-
The valid solution comes from \(\sin x - 1 = 0\), which gives \(\sin x = 1\), leading to \(x = \frac{\pi}{2} + 2n\pi\), where \(n\) is an integer.
-
The steps incorrectly conclude with \(x = \sin^{-1}(3)\) which is invalid because it's outside the range of the sine function.
With this analysis, the main error appears during the factoring of the quadratic equation and incorrectly recognizing potential solutions that don't respect the range of the sine function.
Thus, the option that best describes the solution is:
The equation was factored incorrectly.