1. The problem asks to find the value(s) of $c$ in the interval $[0, \pi]$ such that the average value of the function $f(x) = 4 \sin(x) - 2 \sin(2x)$ equals $f(c)$. 2. We are given the average value $f_{ave} = \frac{8}{\pi}$. 3. To find $c$, solve the equation: $$ 4 \sin(c) - 2 \sin(2c) = \frac{8}{\pi} $$ 4. Use the double-angle identity $\sin(2c) = 2 \sin(c) \cos(c)$: $$ 4 \sin(c) - 2 (2 \sin(c) \cos(c)) = \frac{8}{\pi} $$ $$ 4 \sin(c) - 4 \sin(c) \cos(c) = \frac{8}{\pi} $$ 5. Factor out $4 \sin(c)$: $$ 4 \sin(c) (1 - \cos(c)) = \frac{8}{\pi} $$ 6. Divide both sides by 4: $$ \cancel{4} \sin(c) (1 - \cos(c)) = \frac{8}{\pi} \Rightarrow \sin(c) (1 - \cos(c)) = \frac{8}{4\pi} = \frac{2}{\pi} $$ 7. So we need to solve: $$ \sin(c) (1 - \cos(c)) = \frac{2}{\pi} $$ 8. This transcendental equation can be solved numerically. Using numerical methods or a calculator, the solutions in $[0, \pi]$ are approximately: $$ c \approx 0.927, 2.214 $$ 9. Therefore, the values of $c$ such that $f(c) = f_{ave}$ are: $$ c = 0.927, 2.214 $$