The area of the region whose boundaries are defined by the curves y=2 cos x, y=3 tan x, and the y-axis is

Options

A. `1+3" In "((2)/(sqrt(3)))` sq. units
B. `1+(3)/(2)" In "3-3" In "2` sq. units
C. `1+(3)/(2)" In "3-"In "2` sq. units
D. `" In "3-" In "2` sq. units

Explanation

To find the area of the region bounded by the curves \( y = 2 \cos x \), \( y = 3 \tan x \), and the y-axis, we will follow these steps:

### Step 1: Find the points of intersection
We need to find the points where the curves intersect. This is done by setting the equations equal to each other:

\[ 2 \cos x = 3 \tan x \]

Substituting \(\tan x = \frac{\sin x}{\cos x}\):

\[ 2 \cos x = 3 \frac{\sin x}{\cos x} \]

Multiplying both sides by \(\cos x\) (assuming \(\cos x \neq 0\)):

\[ 2 \cos^2 x = 3 \sin x \]

Using the identity \(\sin^2 x + \cos^2 x = 1\), we can express \(\cos^2 x\) in terms of \(\sin x\):

\[ 2(1 - \sin^2 x) = 3 \sin x \]

This simplifies to:

\[ 2 - 2 \sin^2 x = 3 \sin x \]

Rearranging gives us a quadratic equation:

\[ 2 \sin^2 x + 3 \sin x - 2 = 0 \]

### Step 2: Solve the quadratic equation
Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 2, b = 3, c = -2 \):

\[ \sin x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]
\[ = \frac{-3 \pm \sqrt{9 + 16}}{4} \]
\[ = \frac{-3 \pm 5}{4} \]

Calculating the two possible values:

1. \( \sin x = \frac{2}{4} = \frac{1}{2} \) (valid)
2. \( \sin x = \frac{-8}{4} = -2 \) (not valid)

Thus, we have:

\[ \sin x = \frac{1}{2} \implies x = \frac{\pi}{6} \]

### Step 3: Determine the area
The area \( A \) between the curves from \( x = 0 \) to \( x = \frac{\pi}{6} \) is given by:

\[ A = \int_0^{\frac{\pi}{6}} (2 \cos x - 3 \tan x) \, dx \]

### Step 4: Calculate the integral
Calculating the integral:

1. The integral of \( 2 \cos x \) is \( 2 \sin x \).
2. The integral of \( 3 \tan x \) is \( 3 \ln |\sec x| \).

Thus, we have:

\[ A = \left[ 2 \sin x - 3 \ln |\sec x| \right]_0^{\frac{\pi}{6}} \]

Calculating at the limits:

At \( x = \frac{\pi}{6} \):

\[ = 2 \sin\left(\frac{\pi}{6}\right) - 3 \ln\left|\sec\left(\frac{\pi}{6}\right)\right| \]
\[ = 2 \cdot \frac{1}{2} - 3 \ln\left|\frac{2}{\sqrt{3}}\right| \]
\[ = 1 - 3 \ln 2 + 3 \ln \sqrt{3} \]
\[ = 1 - 3 \ln 2 + \frac{3}{2} \ln 3 \]

At \( x = 0 \):

\[ = 2 \sin(0) - 3 \ln|\sec(0)| = 0 - 3 \ln(1) = 0 \]

Thus, the area \( A \) simplifies to:

\[ A = 1 - 3 \ln 2 + \frac{3}{2} \ln 3 \]

### Final Answer
The area of the region bounded by the curves is:

\[ A = 1 - 3 \ln 2 + \frac{3}{2} \ln 3 \quad \text{(in square units)} \]