The number of solutions of the equation `2 tan^2theta-7sec theta-2 = 0` in the interval `[0, 6pi]` is

Options

A. 8
B. 6
C. 4
D. Zero

Answer

To find the number of solutions of the equation \(2 \tan^2 \theta - 7 \sec \theta - 2 = 0\) in the interval \([0, 6\pi]\), we can follow these steps:

### Step 1: Rewrite the equation using trigonometric identities
We know that:
\[ 1 + \tan^2 \theta = \sec^2 \theta \]
Thus, we can express \(\tan^2 \theta\) in terms of \(\sec^2 \theta\):
\[ \tan^2 \theta = \sec^2 \theta - 1 \]
Substituting this into the original equation gives:
\[ 2(\sec^2 \theta - 1) - 7 \sec \theta - 2 = 0 \]

### Step 2: Simplify the equation
Expanding the equation:
\[ 2 \sec^2 \theta - 2 - 7 \sec \theta - 2 = 0 \]
This simplifies to:
\[ 2 \sec^2 \theta - 7 \sec \theta - 4 = 0 \]

### Step 3: Solve the quadratic equation
Let \(x = \sec \theta\). The equation becomes:
\[ 2x^2 - 7x - 4 = 0 \]
We can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 2\), \(b = -7\), and \(c = -4\).

Calculating the discriminant:
\[ b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot (-4) = 49 + 32 = 81 \]
Now substituting back into the quadratic formula:
\[ x = \frac{7 \pm \sqrt{81}}{4} = \frac{7 \pm 9}{4} \]
This gives us two solutions:
1. \(x = \frac{16}{4} = 4\)
2. \(x = \frac{-2}{4} = -\frac{1}{2}\)

### Step 4: Analyze the solutions for \(\sec \theta\)
The solutions we found are:
1. \(\sec \theta = 4\)
2. \(\sec \theta = -\frac{1}{2}\)

Since \(\sec \theta = \frac{1}{\cos \theta}\), the second solution \(\sec \theta = -\frac{1}{2}\) is not valid because \(\cos \theta\) must be between -1 and 1, and \(\sec \theta\) cannot be less than -1.

### Step 5: Find valid angles for \(\sec \theta = 4\)
Now, we need to find the angles \(\theta\) such that:
\[ \sec \theta = 4 \implies \cos \theta = \frac{1}{4} \]
The cosine function is positive in the first and fourth quadrants. Therefore, the solutions in one cycle \([0, 2\pi]\) are:
1. \(\theta_1 = \cos^{-1}\left(\frac{1}{4}\right)\)
2. \(\theta_2 = 2\pi - \cos^{-1}\left(\frac{1}{4}\right)\)

### Step 6: Count the solutions in the interval \([0, 6\pi]\)
In the interval \([0, 2\pi]\), we have 2 solutions. Since the cosine function has a period of \(2\pi\), we can find the number of solutions in the interval \([0, 6\pi]\) by multiplying the number of solutions in \([0, 2\pi]\) by the number of complete cycles in \([0, 6\pi]\):
\[ \text{Number of complete cycles} = \frac{6\pi}{2\pi} = 3 \]
Thus, the total number of solutions is:
\[ 2 \times 3 = 6 \]

### Final Answer
The number of solutions of the equation \(2 \tan^2 \theta - 7 \sec \theta - 2 = 0\) in the interval \([0, 6\pi]\) is **6**.