1. **State the problem:** Solve the equation $$2 \arcsin(x^2 - 3x + 1) + \pi = 0$$ and find the sum of all values of $x$ that satisfy it. 2. **Isolate the arcsin term:** $$2 \arcsin(x^2 - 3x + 1) = -\pi$$ Divide both sides by 2: $$\arcsin(x^2 - 3x + 1) = -\frac{\pi}{2}$$ 3. **Recall the range and values of arcsin:** The function $\arcsin(y)$ has range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and is defined for $y \in [-1,1]$. 4. **Evaluate the arcsin equation:** $$\arcsin(z) = -\frac{\pi}{2} \implies z = \sin\left(-\frac{\pi}{2}\right) = -1$$ where $z = x^2 - 3x + 1$. 5. **Set up the quadratic equation:** $$x^2 - 3x + 1 = -1$$ Simplify: $$x^2 - 3x + 2 = 0$$ 6. **Solve the quadratic:** Factor: $$(x - 1)(x - 2) = 0$$ So, $$x = 1 \quad \text{or} \quad x = 2$$ 7. **Check domain of arcsin argument:** For $x=1$, $$1^2 - 3(1) + 1 = 1 - 3 + 1 = -1$$ which is in $[-1,1]$. For $x=2$, $$2^2 - 3(2) + 1 = 4 - 6 + 1 = -1$$ which is also in $[-1,1]$. Both values are valid. 8. **Find the sum of all values:** $$1 + 2 = 3$$ **Final answer:** The sum of all values of $x$ satisfying the equation is **3**.