Give step-by-step solution with explanation and final answer: write an expression for the vertical asymptotes of each of the following: ( a ) f ( x ) = 2 sec ( x + pi / 6 ) - 1 ( b ) g ( x ) = cot ( 4 ( x - pi / 3 ) ) ( c ) f ( x ) = - 3 csc ( 1 / 2 x - pi / 2 ) + 2
Question:
Give step-by-step solution with explanation and final answer: write an expression for the vertical asymptotes of each of the following: ( a ) f ( x ) = 2 sec ( x + pi / 6 ) - 1 ( b ) g ( x ) = cot ( 4 ( x - pi / 3 ) ) ( c ) f ( x ) = - 3 csc ( 1 / 2 x - pi / 2 ) + 2
This is AI generated Response.
Asked by: anil
Created at: 2025-09-28 20:49:03
Asked by: anil
Created at: 2025-09-28 20:49:03
Final Answer:
Rendered
Raw Text
Did you like this answer?
# Vertical Asymptotes of Trigonometric Functions
Let’s find the expressions for the **vertical asymptotes** for each function.
---
## (a) \( f(x) = 2\sec(x + \frac{\pi}{6}) - 1 \)
### **Step 1:**
Recall that \(\sec(\theta)\) has vertical asymptotes where \(\cos(\theta) = \).
### **Step 2:**
Set the inside of cosine to zero:
\[
\cos(x + \frac{\pi}{6}) =
\]
### **Step 3:**
The cosine function is zero at:
\[
x + \frac{\pi}{6} = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}
\]
### **Step 4:**
Solve for \(x\):
\[
x = \frac{\pi}{2} - \frac{\pi}{6} + k\pi = \frac{3\pi - \pi}{6} + k\pi = \frac{2\pi}{6} + k\pi = \frac{\pi}{3} + k\pi
\]
### **Final Answer (a):**
\[
\boxed{x = \frac{\pi}{3} + k\pi, \quad k \in \mathbb{Z}}
\]
---
## (b) \( g(x) = \cot(4(x - \frac{\pi}{3})) \)
### **Step 1:**
Recall that \(\cot(\theta)\) has vertical asymptotes where \(\sin(\theta) = \).
### **Step 2:**
Set the inside of sine to zero:
\[
\sin(4(x - \frac{\pi}{3})) =
\]
### **Step 3:**
Sine is zero at integer multiples of \(\pi\):
\[
4(x - \frac{\pi}{3}) = k\pi, \quad k \in \mathbb{Z}
\]
### **Step 4:**
Solve for \(x\):
\[
x - \frac{\pi}{3} = \frac{k\pi}{4} \implies x = \frac{\pi}{3} + \frac{k\pi}{4}
\]
### **Final Answer (b):**
\[
\boxed{x = \frac{\pi}{3} + \frac{k\pi}{4}, \quad k \in \bb{Z}}
\]
---
## (c) \( f(x) = -3\csc(\frac{}{2} x - \frac{\pi}{2}) + 2 \)
### **Step 1:**
Recall that \(\csc(\theta)\) has vertical asymptotes where \(\sin(\theta) = \).
### **Step 2:**
Set the argument to zeros of sine:
\[
\sin\left(\frac{1}{2}x - \frac{\pi}{2}\right) =
\]
### **Step 3:**
\[
\frac{1}{2}x - \frac{\pi}{2} = k\pi, \quad k \in \mathbb{Z}
\]
### **Step 4:**
Solve for \(x\):
\[
\frac{1}{2}x = k\pi + \frac{\pi}{2}
\]
\[
x = 2k\pi + \pi
\]
### **Final Answer (c):**
\[
\boxed{x = \pi + 2k\pi, \quad k \in \mathbb{Z}}
\]
---
## **Summary Table**
| Function | Vertical Asymptotes |
|-----------------------------------------------------|-----------------------------|
| \( 2\sec(x+\frac{\pi}{6})-1 \) | \( x = \frac{\pi}{3} + k\pi \) |
| \( \cot(4(x-\frac{\pi}{3})) \) | \( x = \frac{\pi}{3} + \frac{k\pi}{4} \)|
| \( -3\csc(\frac{1}{2}x-\frac{\pi}{2})+2 \) | \( x = \pi + 2k\pi \) |
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :
VIPSolutions