The number of roots of the equation `3 sin^(2)x =8 cos x" in "(-(pi)/(2), (pi)/(2))` is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equation \( 3 \sin^2 x = 8 \cos x \) in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that \( \sin^2 x = 1 - \cos^2 x \). We can substitute this into the equation:

\[ 3 \sin^2 x = 8 \cos x \implies 3(1 - \cos^2 x) = 8 \cos x \]

### Step 2: Rearrange the equation
Distributing the 3 gives us:

\[ 3 - 3 \cos^2 x = 8 \cos x \]

Now, rearranging this equation leads to:

\[ 3 \cos^2 x + 8 \cos x - 3 = 0 \]

### Step 3: Form a quadratic equation
This is a standard quadratic equation in terms of \( \cos x \). We can denote \( y = \cos x \), so the equation becomes:

\[ 3y^2 + 8y - 3 = 0 \]

### Step 4: Solve the quadratic equation
We can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 3 \), \( b = 8 \), and \( c = -3 \).

Calculating the discriminant:

\[ b^2 - 4ac = 8^2 - 4 \cdot 3 \cdot (-3) = 64 + 36 = 100 \]

Now applying the quadratic formula:

\[ y = \frac{-8 \pm \sqrt{100}}{2 \cdot 3} = \frac{-8 \pm 10}{6} \]

Calculating the two potential solutions:

1. \( y_1 = \frac{2}{6} = \frac{1}{3} \)
2. \( y_2 = \frac{-18}{6} = -3 \)

### Step 5: Analyze the solutions
The first solution \( y_1 = \frac{1}{3} \) is valid since \( \cos x \) can take values between -1 and 1.

The second solution \( y_2 = -3 \) is not valid because the range of the cosine function is \([-1, 1]\).

### Step 6: Find the angles corresponding to the valid solution
For \( \cos x = \frac{1}{3} \):

\[ x = \cos^{-1}\left(\frac{1}{3}\right) \]

Since we are looking for solutions in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), the cosine function is positive in this interval, and we have:

\[ x = \cos^{-1}\left(\frac{1}{3}\right) \quad \text{and} \quad x = -\cos^{-1}\left(\frac{1}{3}\right) \]

### Conclusion
Thus, there are two solutions in the specified interval:

1. \( x = \cos^{-1}\left(\frac{1}{3}\right) \)
2. \( x = -\cos^{-1}\left(\frac{1}{3}\right) \)

The number of roots of the equation \( 3 \sin^2 x = 8 \cos x \) in the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) is **2**.

---