Let a, b, c be three non-zero real numbers such that the equation √3a cos x + 2b sin x = c, x ∈ (-π/

Q: Wait, why can't we just solve for $x$ directly using the quadratic formula?

That is a great instinct, but notice we do not know the values of $a, b,$ or $c$. We are dealing with a general equation. Instead of solving for $x$ algebraically, we use the properties of the function's graph—specifically its symmetry—to relate the roots to the parameters.

Q: How do we know for sure that $\alpha + \beta = 2\theta$?

Excellent question! The function $y = \cos(x - \theta)$ is symmetric about the vertical line $x = \theta$. If a horizontal line $y = k$ intersects this curve at two points, those points must be equidistant from the axis of symmetry. Thus, the midpoint of the roots is exactly $\theta$.

Q: Is it possible for the roots to be outside the interval $(-\pi/2, \pi/2)$?

The problem statement restricts the roots to this interval. In a competitive exam setting, we trust the constraints given. The symmetry argument holds regardless of the interval, provided the roots exist within it.

Question

Let $a, b, c$ be three non-zero real numbers such that the equation $\sqrt{3}a \cos x + 2b \sin x = c, x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ has two distinct real roots $\alpha$ and $\beta$ with $\alpha + \beta = \frac{\pi}{3}$. Then, the value of $\frac{b}{a}$ is ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the equation:

$$\sqrt{3}a \cos x + 2b \sin x = c$$

This equation has two distinct roots, $\alpha$ and $\beta$, such that $\alpha + \beta = \frac{\pi}{3}$. Our objective is to determine the ratio $\frac{b}{a}$.

When encountering an expression of the form $A \cos x + B \sin x$, the **auxiliary angle method** is the most efficient tool. We define a scaling factor $R$ as:

$$R = \sqrt{(\sqrt{3}a)^2 + (2b)^2}$$

Dividing the entire equation by $R$, we obtain:

$$\frac{\sqrt{3}a}{R} \cos x + \frac{2b}{R} \sin x = \frac{c}{R}$$

Since the sum of the squares of the new coefficients is unity, we define an angle $	heta$ such that $\cos 	heta = \frac{\sqrt{3}a}{R}$ and $\sin 	heta = \frac{2b}{R}$. This yields the relation:

$$	an 	heta = \frac{2b}{\sqrt{3}a}$$

### The Geometry of Symmetry

The equation now simplifies to:

$$\cos x \cos 	heta + \sin x \sin 	heta = \frac{c}{R}$$

This is equivalent to the compact form:

$$\cos(x - 	heta) = \frac{c}{R}$$

The roots $\alpha$ and $\beta$ represent the $x$-coordinates where the wave $y = \cos(x - 	heta)$ intersects the horizontal line $y = \frac{c}{R}$. Because the cosine function is symmetric about its peak, the curve is symmetric about the vertical line $x = 	heta$.

Any horizontal line intersecting the curve at two points must do so at points equidistant from the axis of symmetry. Therefore, $	heta$ is the arithmetic mean of the roots:

$$	heta = \frac{\alpha + \beta}{2}$$

### The Final Resolution

Given that $\alpha + \beta = \frac{\pi}{3}$, we substitute this into our midpoint relation:

$$	heta = \frac{\pi/3}{2} = \frac{\pi}{6}$$

Returning to our algebraic bridge, we substitute $	heta = \frac{\pi}{6}$ into the expression for $	an 	heta$:

$$	an\left(\frac{\pi}{6}ight) = \frac{2b}{\sqrt{3}a}$$

Since $	an\left(\frac{\pi}{6}ight) = \frac{1}{\sqrt{3}}$, the equation becomes:

$$\frac{1}{\sqrt{3}} = \frac{2b}{\sqrt{3}a}$$

Canceling $\sqrt{3}$ from both denominators, we arrive at $1 = \frac{2b}{a}$. Thus, the final ratio is:

$$\mathbf{\frac{b}{a} = \frac{1}{2} = 0.5}$$

Enter Numerical Value:

Visualized Solution

The Given Equation

Auxiliary Angle Method

Defining the Angle $θ$

Condensing the Equation

Graphing the Equation

Locating the Roots

Symmetry of the Cosine Wave

Midpoint of the Roots

Calculating $θ$

Connecting $θ$ to $a$ and $b$

Evaluating the Tangent

Solving for the Ratio

Final Answer

The Sigma Insight: General Solution of Trigonometric Equations

Analyzing the Setup

The Geometry of Symmetry

The Final Resolution

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