Question:

Step 1: Recognize the telescoping pattern using a key trigonometric identity. We will use the identity: $$\frac{\sin A}{\cos 3A} = \frac{1}{2}(\tan 3A - \tan A)$$ This identity allows us to rewrite each term in the given equation in a form that will telescope (cancel out intermediate terms). Step 2: Apply the identity to each term in the equation. For the first term: $$\frac{\sin x}{\cos 3x} = \frac{1}{2}(\tan 3x - \tan x)$$ For the second term: $$\frac{\sin 3x}{\cos 9x} = \frac{1}{2}(\tan 9x - \tan 3x)$$ For the third term: $$\frac{\sin 9x}{\cos 27x} = \frac{1}{2}(\tan 27x - \tan 9x)$$ Step 3: Add all three terms together. $$\frac{\sin x}{\cos 3x} + \frac{\sin 3x}{\cos 9x} + \frac{\sin 9x}{\cos 27x} = \frac{1}{2}(\tan 3x - \tan x) + \frac{1}{2}(\tan 9x - \tan 3x) + \frac{1}{2}(\tan 27x - \tan 9x)$$ Step 4: Simplify by observing the telescoping cancellation. When we expand and combine the terms: $$= \frac{1}{2}[\tan 3x - \tan x + \tan 9x - \tan 3x + \tan 27x - \tan 9x]$$ Notice that $\tan 3x$ cancels with $-\tan 3x$, and $\tan 9x$ cancels with $-\tan 9x$, leaving only: $$= \frac{1}{2}(\tan 27x - \tan x)$$ Step 5: Set the sum equal to zero and solve. Given that the original equation equals zero: $$\frac{1}{2}(\tan 27x - \tan x) = 0$$ Therefore: $$\tan 27x - \tan x = 0$$ $$\tan 27x = \tan x$$ Step 6: Use the general solution for the tangent equation. The general solution for $\tan A = \tan B$ is $A = B + n\pi$ where $n$ is any integer. Therefore: $$27x = x + n\pi$$ Step 7: Solve for $x$. $$27x - x = n\pi$$ $$26x = n\pi$$ $$x = \frac{n\pi}{26}$$ Step 8: Find the least positive value. For the least positive value of $x$, we use the smallest positive integer value of $n$, which is $n = 1$: $$x = \frac{\pi}{26}$$ **Final Answer:** The least positive value of $x$ satisfying the given equation is $\boxed{\frac{\pi}{26}}$, which corresponds to **Option 1**. <div class="key-concept"><strong>Key Concept:</strong> Telescoping trigonometric sums using the identity $\dfrac{\sin A}{\cos 3A} = \dfrac{1}{2}(\tan 3A - \tan A)$</div> <div class="trap-box"><strong>Trap:</strong> The telescoping identity reduces the sum to $\tan 27x - \tan x = 0$, giving $x = \dfrac{n\pi}{26}$, not $\dfrac{n\pi}{27}$.</div>