Question:

Step 1: Simplify the given constraint equation. From the given condition $\cos x + \cos^2 x = 1$, we can rearrange to get: $$\cos x = 1 - \cos^2 x = \sin^2 x$$ This is a key relationship that will help us evaluate $E$. Step 2: Recognize the binomial expansion pattern in $E$. We observe that the expression $E = \sin^{12} x + 3\sin^{10} x + 3\sin^8 x + \sin^6 x + 2$ contains coefficients $1, 3, 3, 1$ which match the binomial expansion of $(a+b)^3$. We can rewrite: $$E = (\sin^4 x + \sin^2 x)^3 + 2$$ This is because $(\sin^4 x + \sin^2 x)^3 = \sin^{12}x + 3\sin^{10}x + 3\sin^8x + \sin^6x$. Step 3: Express $\sin^4 x + \sin^2 x$ in terms of $\cos x$. Using the relationship $\sin^2 x = \cos x$ from Step 1, we have $\sin^4 x = \cos^2 x$. Therefore: $$\sin^4 x + \sin^2 x = \cos^2 x + \cos x = \cos x(\cos x + 1)$$ Step 4: Use the constraint to evaluate the cubic term. From the given condition $\cos x + \cos^2 x = 1$, we can factor: $$\cos x(1 + \cos x) = 1$$ Therefore: $$(\sin^4 x + \sin^2 x)^3 = [\cos x(\cos x + 1)]^3 = 1^3 = 1$$ Step 5: Calculate the value of $E$. Substituting back into our expression for $E$: $$E = 1 + 2 = 3$$ Step 6: Evaluate the logarithm. We need to find $\log_{\tan\frac{\pi}{3}} E$. First, recall that $\tan\dfrac{\pi}{3} = \sqrt{3}$. Therefore: $$\log_{\sqrt{3}} 3 = \log_{\sqrt{3}} (\sqrt{3})^2 = 2$$ **Final Answer:** The value of $\log_{\tan\frac{\pi}{3}} E = 2$ The correct option is **Option 2: 2**