1. **State the problem:** We have a right triangle ABC with $\angle A = 90^\circ$. Given $\sin B = 6x - 0.67$ and $\cos C = 3x - 0.76$, we need to solve for $x$. 2. **Recall important facts:** In a right triangle, the two acute angles $B$ and $C$ satisfy $B + C = 90^\circ$. 3. **Use the complementary angle identity:** Since $C = 90^\circ - B$, we have $\cos C = \cos(90^\circ - B) = \sin B$. 4. **Set the expressions equal:** $$ \cos C = \sin B \implies 3x - 0.76 = 6x - 0.67 $$ 5. **Solve for $x$:** $$ 3x - 0.76 = 6x - 0.67 $$ Subtract $3x$ from both sides: $$ \cancel{3x} - 0.76 = \cancel{3x} + 3x - 0.67 \implies -0.76 = 3x - 0.67 $$ Add $0.67$ to both sides: $$ -0.76 + 0.67 = 3x - \cancel{0.67} + \cancel{0.67} \implies -0.09 = 3x $$ Divide both sides by 3: $$ \frac{-0.09}{\cancel{3}} = \frac{3x}{\cancel{3}} \implies x = -0.03 $$ 6. **Final answer:** $x = -0.03$.