1. Interpret the given expression

We are given $e^{\left(\cos^2 x+\cos^4 x+\cos^6 x+\cdots\right)\log_e 2}$ Let $S=\cos^2 x+\cos^4 x+\cos^6 x+\cdots$ Then the expression becomes $e^{S\ln 2}=2^S$ So if we denote this by $t$, then $t=2^S$

It is given that $t$ satisfies $t^2-9t+8=0$ Factorizing, $t^2-9t+8=(t-1)(t-8)=0$ Hence $t=1 \quad \text{or} \quad t=8$

2. Find possible values of $S$

Since $t=2^S$ we get:

• If $t=1$, then $2^S=1 \Rightarrow S=0$
• If $t=8$, then $2^S=8=2^3 \Rightarrow S=3$

Now, $S=\cos^2 x+\cos^4 x+\cos^6 x+\cdots$ This is a geometric series with first term $\cos^2 x$ and common ratio $\cos^2 x$. So $S=\frac{\cos^2 x}{1-\cos^2 x}=\frac{\cos^2 x}{\sin^2 x}=\cot^2 x$ (valid since $0<x<\pi/2$, so $\sin x\neq 0$).

Thus $\cot^2 x=0 \quad \text{or} \quad \cot^2 x=3$

But for $0<x<\pi/2$, we have $\cot x>0$, so:

• $\cot^2 x=0$ would imply $\cot x=0$, i.e. $x=\pi/2$, not allowed.
• Therefore the valid case is $\cot^2 x=3 \Rightarrow \cot x=\sqrt{3}$ Hence $\tan x=\frac{1}{\sqrt3}$ So in the first quadrant, $x=\frac{\pi}{6}$

3. Evaluate the required expression

We need $\frac{2\sin x}{\sin x+\sqrt3\cos x}$ At $x=\pi/6$, $\sin\frac\pi6=\frac12, \qquad \cos\frac\pi6=\frac{\sqrt3}{2}$ Thus