The least difference between the roots, in the first quadrant `(0 <= x <= pi/2),` of the equation `4 cos x(2-3 sin^2 x)+(cos 2x+1)=0, `is

Options

A. equal to ` (pi)/(2)`
B. ` gt (pi)/(10) `
C. ` lt (pi)/(2) `
D. ` lt (pi)/(3) `

Answer

To solve the equation \( 4 \cos x (2 - 3 \sin^2 x) + (\cos 2x + 1) = 0 \) and find the least difference between the roots in the first quadrant, we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 4 \cos x (2 - 3 \sin^2 x) + (\cos 2x + 1) = 0 \]
Using the identity \( \cos 2x = 2 \cos^2 x - 1 \), we can substitute for \( \cos 2x \):
\[ 4 \cos x (2 - 3 \sin^2 x) + (2 \cos^2 x - 1 + 1) = 0 \]
This simplifies to:
\[ 4 \cos x (2 - 3 \sin^2 x) + 2 \cos^2 x = 0 \]

### Step 2: Substitute \(\sin^2 x\)
Recall that \( \sin^2 x = 1 - \cos^2 x \). Substituting this into the equation gives:
\[ 4 \cos x \left(2 - 3(1 - \cos^2 x)\right) + 2 \cos^2 x = 0 \]
This simplifies to:
\[ 4 \cos x (2 - 3 + 3 \cos^2 x) + 2 \cos^2 x = 0 \]
\[ 4 \cos x (3 \cos^2 x - 1) + 2 \cos^2 x = 0 \]

### Step 3: Factor the equation
We can factor out \( 2 \cos x \):
\[ 2 \cos x (2(3 \cos^2 x - 1) + \cos^2 x) = 0 \]
This leads to:
\[ 2 \cos x (6 \cos^2 x - 2 + \cos^2 x) = 0 \]
\[ 2 \cos x (7 \cos^2 x - 2) = 0 \]

### Step 4: Solve for roots
Setting \( 2 \cos x = 0 \) gives:
\[ \cos x = 0 \quad \Rightarrow \quad x = \frac{\pi}{2} \]
Setting \( 7 \cos^2 x - 2 = 0 \) gives:
\[ \cos^2 x = \frac{2}{7} \quad \Rightarrow \quad \cos x = \pm \sqrt{\frac{2}{7}} \]
Since we are in the first quadrant, we take:
\[ \cos x = \sqrt{\frac{2}{7}} \quad \Rightarrow \quad x = \cos^{-1}\left(\sqrt{\frac{2}{7}}\right) \]

### Step 5: Find the angles
Now we have two angles:
1. \( x_1 = \frac{\pi}{2} \)
2. \( x_2 = \cos^{-1}\left(\sqrt{\frac{2}{7}}\right) \)

### Step 6: Calculate the least difference
To find the least difference between the roots:
\[ \text{Difference} = x_1 - x_2 = \frac{\pi}{2} - \cos^{-1}\left(\sqrt{\frac{2}{7}}\right) \]
Using the identity \( \cos^{-1}(a) + \sin^{-1}(a) = \frac{\pi}{2} \):
\[ \frac{\pi}{2} - \cos^{-1}\left(\sqrt{\frac{2}{7}}\right) = \sin^{-1}\left(\sqrt{\frac{2}{7}}\right) \]

### Step 7: Final answer
The least difference between the roots in the first quadrant is:
\[ \sin^{-1}\left(\sqrt{\frac{2}{7}}\right) \]