Question:

Step 1: Set up the equation using substitution. Let $A = \cos^{-1}\left(\frac{2}{3x}\right)$ and $B = \cos^{-1}\left(\frac{3}{4x}\right)$. Given that $A + B = \frac{\pi}{2}$, we can write: $$B = \frac{\pi}{2} - A$$ Step 2: Use the complementary angle relationship. Taking cosine of both sides of $B = \frac{\pi}{2} - A$: $$\cos B = \cos\left(\frac{\pi}{2} - A\right) = \sin A$$ From the definitions of $A$ and $B$, we know: $$\cos B = \frac{3}{4x}$$ Step 3: Express $\sin A$ in terms of $\cos A$. Since $A = \cos^{-1}\left(\frac{2}{3x}\right)$, we have $\cos A = \frac{2}{3x}$. Using the Pythagorean identity: $$\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \frac{4}{9x^2}}$$ Step 4: Set up the equation from the complementary angle relationship. From Step 2, we have $\cos B = \sin A$: $$\frac{3}{4x} = \sqrt{1 - \frac{4}{9x^2}}$$ Step 5: Square both sides to eliminate the square root. $$\left(\frac{3}{4x}\right)^2 = 1 - \frac{4}{9x^2}$$ $$\frac{9}{16x^2} = 1 - \frac{4}{9x^2}$$ Step 6: Clear the denominators by multiplying through by $144x^2$. The LCD of $16x^2$ and $9x^2$ is $144x^2$. Multiplying the entire equation: $$144x^2 \cdot \frac{9}{16x^2} = 144x^2 \cdot 1 - 144x^2 \cdot \frac{4}{9x^2}$$ $$81 = 144x^2 - 64$$ Step 7: Solve for $x^2$. $$144x^2 = 81 + 64$$ $$144x^2 = 145$$ $$x^2 = \frac{145}{144}$$ Step 8: Find $x$ and verify the domain condition. Taking the positive square root (since $x > \frac{3}{4}$): $$x = \frac{\sqrt{145}}{12}$$ We can verify: $\frac{\sqrt{145}}{12} \approx \frac{12.04}{12} \approx 1.003 > \frac{3}{4}$ ✓ **Final Answer:** $x = \dfrac{\sqrt{145}}{12}$ This corresponds to **Option 4** (or Option 1 if the numbering in the problem statement is different).