Question: sin x^2 = 0.81
1. **State the problem:** Solve the equation $\sin(x^2) = 0.81$ for $x$.
2. **Recall the sine function properties:** The sine function satisfies $\sin \theta = y$ where $\theta = x^2$ in this case.
3. **Find the general solutions for $\theta$:** Since $\sin \theta = 0.81$, the principal solution is $\theta = \arcsin(0.81)$.
4. **Calculate $\arcsin(0.81)$:**
$$\arcsin(0.81) \approx 0.944$$ radians.
5. **General solutions for sine:**
$$\theta = 0.944 + 2k\pi \quad \text{or} \quad \theta = \pi - 0.944 + 2k\pi$$
where $k$ is any integer.
6. **Substitute back $\theta = x^2$:**
$$x^2 = 0.944 + 2k\pi \quad \text{or} \quad x^2 = \pi - 0.944 + 2k\pi$$
7. **Solve for $x$:**
$$x = \pm \sqrt{0.944 + 2k\pi} \quad \text{or} \quad x = \pm \sqrt{2.198 + 2k\pi}$$
8. **Summary:** The solutions are all real numbers $x$ such that
$$x = \pm \sqrt{0.944 + 2k\pi}$$
or
$$x = \pm \sqrt{2.198 + 2k\pi}$$
for any integer $k$.
This gives infinitely many solutions because sine is periodic.
**Final answer:**
$$x = \pm \sqrt{0.944 + 2k\pi} \quad \text{or} \quad x = \pm \sqrt{2.198 + 2k\pi}, \quad k \in \mathbb{Z}$$