Find the number of roots of the equation `16 sec^(3) theta - 12 tan^(2) theta - 4 sec theta =9` in interval `(-pi,pi)`

Answer

To solve the equation \( 16 \sec^3 \theta - 12 \tan^2 \theta - 4 \sec \theta = 9 \) in the interval \( (-\pi, \pi) \), we will follow these steps:

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

\[ 16 \sec^3 \theta - 12 (\sec^2 \theta - 1) - 4 \sec \theta = 9 \]

### Step 2: Simplify the equation
Now, let's simplify the equation:

\[ 16 \sec^3 \theta - 12 \sec^2 \theta + 12 - 4 \sec \theta - 9 = 0 \]

This simplifies to:

\[ 16 \sec^3 \theta - 12 \sec^2 \theta - 4 \sec \theta + 3 = 0 \]

### Step 3: Factor the polynomial
We can factor out common terms. Let's rearrange the equation:

\[ 16 \sec^3 \theta - 12 \sec^2 \theta - 4 \sec \theta + 3 = 0 \]

We can try to factor this polynomial. We can take \( 4 \sec^2 \theta \) common from the first two terms:

\[ 4 \sec^2 \theta (4 \sec \theta - 3) - 1(4 \sec \theta - 3) = 0 \]

This can be factored as:

\[ (4 \sec^2 \theta - 1)(4 \sec \theta - 3) = 0 \]

### Step 4: Solve the factored equations
Now we have two equations to solve:

1. \( 4 \sec^2 \theta - 1 = 0 \)
2. \( 4 \sec \theta - 3 = 0 \)

#### For the first equation:
\[ 4 \sec^2 \theta = 1 \implies \sec^2 \theta = \frac{1}{4} \implies \sec \theta = \pm \frac{1}{2} \]
However, the secant function is defined as \( \sec \theta = \frac{1}{\cos \theta} \), and since \( \sec \theta \) cannot take values between -1 and 1, there are no solutions from this equation.

#### For the second equation:
\[ 4 \sec \theta = 3 \implies \sec \theta = \frac{3}{4} \]
Again, since \( \sec \theta \) must be greater than or equal to 1 or less than or equal to -1, this value is also not valid.

### Step 5: Conclusion
Since neither of the equations provides valid solutions for \( \sec \theta \), we conclude that there are no roots for the original equation in the interval \( (-\pi, \pi) \).

### Final Answer
The number of roots of the equation \( 16 \sec^3 \theta - 12 \tan^2 \theta - 4 \sec \theta = 9 \) in the interval \( (-\pi, \pi) \) is **0**.

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