Question

If $e^{(\cos^2 x + \cos^4 x + \cos^6 x + .......) \log_e 2}$ satisfies $t^2 - 9t + 8 = 0$, then find $\frac{2 \sin x}{\sin x + \sqrt{3} \cos x}$ for $0<x<\pi/2$ :

• A. $1/2$
• B. $\sqrt{3}$
• C. $3/2$
• D. $2\sqrt{3}$

Hint:

For infinite geometric series in exponents, always simplify the series first. $a^x$ is equivalent to $e^{x \ln a}$.

Answer 1

The Correct Option is A

Step 1: Solve the exponent GP: $S_{\infty} = \frac{\cos^2 x}{1 - \cos^2 x} = \cot^2 x$.
Step 2: The term is $e^{(\cot^2 x) \ln 2} = 2^{\cot^2 x}$.
Step 3: Roots of $t^2 - 9t + 8 = 0$ are 1 and 8.
Step 4: $2^{\cot^2 x} = 8 \Rightarrow \cot^2 x = 3 \Rightarrow \cot x = \sqrt{3} \Rightarrow x = 30^\circ$.
Step 5: $\frac{2 \sin 30^\circ}{\sin 30^\circ + \sqrt{3} \cos 30^\circ} = \frac{2(1/2)}{1/2 + \sqrt{3}(\sqrt{3}/2)} = \frac{1}{2}$.