If e^((cos² x + cos⁴ x + cos⁶ x + ... ∞) log_e 2) satisfies the equation t² - 9t + 8 = 0, then the v

Q: Wait, why can't we just use $\sin^2 x + \cos^2 x = 1$ immediately on the series?

That is a great instinct! However, the identity $\sin^2 x + \cos^2 x = 1$ applies to the *sum* of the series, not the individual terms. We must first sum the infinite geometric progression to get $\frac{\cos^2 x}{1 - \cos^2 x}$, and *then* use the identity to simplify the denominator into $\sin^2 x$.

Q: Why do we reject the root $t=1$?

Excellent catch. We found $t = 2^{\cot^2 x}$. If $t=1$, then $2^{\cot^2 x} = 1$, which means $\cot^2 x = 0$, so $\cot x = 0$. This implies $x = \frac{\pi}{2}$. But the problem explicitly states $0 < x < \frac{\pi}{2}$. Therefore, $x$ cannot be $\frac{\pi}{2}$, and we must reject $t=1$.

Q: Is there a faster way than finding $\sin x$ and $\cos x$ individually?

Absolutely! Instead of calculating $\sin x$ and $\cos x$ separately, divide both the numerator and the denominator of the target expression by $\sin x$. This transforms the expression into a function of $\cot x$ alone, which we already know is $\sqrt{3}$. It saves time and reduces the chance of calculation errors.

Question

If $e^{(\cos^2 x + \cos^4 x + \cos^6 x + \dots \infty) \log_e 2}$ satisfies the equation $t^2 - 9t + 8 = 0$, then the value of $\frac{2 \sin x}{\sin x + \sqrt{3} \cos x}$ $(0 < x < \frac{\pi}{2})$ is

Options:

(A) $\frac{3}{2}$

(B) $2\sqrt{3}$

(C) $\frac{1}{2}$

(D) $\sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are staring at a complex JEE Advanced problem. You see an expression like $e^{(\cos^2 x + \cos^4 x + \cos^6 x + \dots \infty) \ln 2}$.

It looks like a mathematical behemoth, designed to intimidate you. However, the secret of the JEE is that complexity is often just a mask for simplicity.

### Taming the Infinite Series

First, let us isolate the exponent, which is an infinite series: $\cos^2 x + \cos^4 x + \cos^6 x + \dots \infty$.

Each term is the previous term multiplied by $\cos^2 x$. This is a classic infinite Geometric Progression (GP) where the first term $a = \cos^2 x$ and the common ratio $r = \cos^2 x$.

Since $0 < x < \frac{\pi}{2}$, we know that $0 < \cos^2 x < 1$, ensuring the series converges. The sum of an infinite GP is given by $S = \frac{a}{1-r}$.

Substituting our values:

$$S = \frac{\cos^2 x}{1 - \cos^2 x}$$

Using the fundamental trigonometric identity $\sin^2 x + \cos^2 x = 1$, we know $1 - \cos^2 x = \sin^2 x$. Thus, the sum simplifies beautifully:

$$S = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x$$

### The Logarithmic Bridge

Now, we substitute this back into the original expression: $e^{\cot^2 x \cdot \ln 2}$.

Using the logarithmic property $k \ln a = \ln(a^k)$, this becomes $e^{\ln(2^{\cot^2 x})}$.

Since the exponential function $e^x$ and the natural logarithm $\ln x$ are inverse functions, they cancel each other out:

$$e^{\ln(2^{\cot^2 x})} = 2^{\cot^2 x}$$

### The Quadratic Gatekeeper

The problem states that this expression satisfies the quadratic equation $t^2 - 9t + 8 = 0$, where $t = 2^{\cot^2 x}$.

Factoring the quadratic:

$$t^2 - 9t + 8 = (t - 1)(t - 8) = 0$$

This yields two potential roots: $t = 1$ or $t = 8$.

If $t = 1$, then $2^{\cot^2 x} = 1$, which implies $\cot^2 x = 0$, or $\cot x = 0$. This results in $x = \frac{\pi}{2}$, which violates our constraint $0 < x < \frac{\pi}{2}$.

Therefore, we must have $t = 8$:

$$2^{\cot^2 x} = 8 = 2^3$$

This implies $\cot^2 x = 3$, or $\cot x = \sqrt{3}$ (taking the positive root as $x$ is in the first quadrant).

### The Trigonometric Finale

We need to evaluate the expression $E = \frac{2 \sin x}{\sin x + \sqrt{3} \cos x}$.

To simplify, divide both the numerator and the denominator by $\sin x$:

$$E = \frac{2}{1 + \sqrt{3} \cot x}$$

Now, substitute $\cot x = \sqrt{3}$ into the equation:

$$E = \frac{2}{1 + \sqrt{3}(\sqrt{3})} = \frac{2}{1 + 3} = \frac{2}{4} = \frac{1}{2}$$

The final answer is **$\frac{1}{2}$**.

Animated Solution for Mathematics - Trigonometry: If $e^{(c o s^{2} x + c o s^{4} x + c o s^{6} x + \dots \infty) l o g_{e} 2}$ satisfies the equation $t^{2} - 9 t + 8 = 0$ , then the value of $\frac{2 s i n x}{s i n x + 3 c o s x}$ $(0 < x < \frac{π}{2})$ is

Select Answer:

Visualized Solution

The Exponential Behemoth

Identifying the Geometric Progression

Sum of Infinite GP

Simplifying the Exponent

Condensing the Base Expression

The Quadratic Connection

Solving for $t$

Evaluating the Roots

Finding $cot x$

Geometric Interpretation

The Target Expression

Smart Algebraic Manipulation

Substituting the Value

The Final Answer

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

Taming the Infinite Series

The Logarithmic Bridge

The Quadratic Gatekeeper

The Trigonometric Finale

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