For ` a in [pi , 2 pi]` and `n in Z` the critical points of g
`f(x) = 1/3 sin a tan ^3 "" x + (sin a-1)tan x + sqrt(a-2)/(8-a)` are

Options

A. `x = n pi`
B. `x=2n pi`
C. `x= (2n +1)pi`
D. none of these

Explanation

To find the critical points of the function

\[ f(x) = \frac{1}{3} \sin a \tan^3 x + (\sin a - 1) \tan x + \frac{\sqrt{a - 2}}{8 - a} \]

for \( a \in [\pi, 2\pi] \) and \( n \in \mathbb{Z} \), we need to follow these steps:

### Step 1: Differentiate the Function
We start by differentiating \( f(x) \) with respect to \( x \):

\[ f'(x) = \frac{d}{dx} \left( \frac{1}{3} \sin a \tan^3 x + (\sin a - 1) \tan x + \frac{\sqrt{a - 2}}{8 - a} \right) \]

Using the chain rule and the derivative of \( \tan x \):

\[ f'(x) = \frac{1}{3} \sin a \cdot 3 \tan^2 x \cdot \sec^2 x + (\sin a - 1) \sec^2 x \]

Simplifying this, we get:

\[ f'(x) = \sin a \tan^2 x \sec^2 x + (\sin a - 1) \sec^2 x \]

### Step 2: Set the Derivative to Zero
To find the critical points, we set the derivative equal to zero:

\[ \sin a \tan^2 x \sec^2 x + (\sin a - 1) \sec^2 x = 0 \]

Factoring out \( \sec^2 x \):

\[ \sec^2 x \left( \sin a \tan^2 x + (\sin a - 1) \right) = 0 \]

Since \( \sec^2 x \) is never zero, we focus on:

\[ \sin a \tan^2 x + (\sin a - 1) = 0 \]

### Step 3: Solve for \( \tan^2 x \)
Rearranging gives us:

\[ \sin a \tan^2 x = 1 - \sin a \]

Thus,

\[ \tan^2 x = \frac{1 - \sin a}{\sin a} \]

### Step 4: Analyze the Result
For \( \tan^2 x \) to be non-negative, we require:

\[ 1 - \sin a \geq 0 \quad \text{and} \quad \sin a > 0 \]

This means:

1. \( \sin a \leq 1 \)
2. \( \sin a > 0 \)

Given \( a \in [\pi, 2\pi] \), \( \sin a \) will be negative, which means \( 1 - \sin a \) will always be positive. Therefore, \( \tan^2 x \) will always be non-negative.

### Step 5: Conclusion
The critical points occur where \( \tan^2 x = \frac{1 - \sin a}{\sin a} \). Since \( \tan x \) is periodic, we can find multiple solutions for \( x \) in the interval \( [\pi, 2\pi] \).