1. **State the problem:** Solve the equation $$\sin^2 x - 5 \sin x + 6 = 2$$ for $x$. 2. **Rewrite the equation:** Move all terms to one side to set the equation equal to zero: $$\sin^2 x - 5 \sin x + 6 - 2 = 0$$ which simplifies to $$\sin^2 x - 5 \sin x + 4 = 0$$ 3. **Use substitution:** Let $y = \sin x$. The equation becomes $$y^2 - 5y + 4 = 0$$ 4. **Factor the quadratic:** $$y^2 - 5y + 4 = (y - 4)(y - 1) = 0$$ 5. **Solve for $y$:** $$y - 4 = 0 \implies y = 4$$ $$y - 1 = 0 \implies y = 1$$ 6. **Check the domain of $\sin x$:** Since $\sin x$ must be between $-1$ and $1$, $y=4$ is not possible. 7. **Solve for $x$ when $\sin x = 1$:** $$\sin x = 1$$ This occurs at $$x = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}$$ **Final answer:** $$x = \frac{\pi}{2} + 2k\pi, \quad k \in \mathbb{Z}$$