Answer :
Sure! Let's solve the equation [tex]\(\sec(x) \cos(3x) = 0\)[/tex] within the interval [tex]\(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)[/tex].
### Step 1: Understanding the Functions Involved
The given equation involves the secant function [tex]\(\sec(x)\)[/tex] and the cosine function [tex]\(\cos(3x)\)[/tex]. Recall that:
- [tex]\(\sec(x) = \frac{1}{\cos(x)}\)[/tex]
- [tex]\(\cos(kx)\)[/tex] is the cosine function with argument [tex]\(kx\)[/tex], where [tex]\(k\)[/tex] is a constant.
### Step 2: Setting Up the Equation
The equation is:
[tex]\[ \sec(x) \cos(3x) = 0 \][/tex]
For the product of two functions to be zero, at least one of the functions must be zero.
### Step 3: Solving Each Part
Let's solve for each part separately.
#### Part A: When [tex]\(\sec(x) = 0\)[/tex]
[tex]\(\sec(x)\)[/tex] is the reciprocal of [tex]\(\cos(x)\)[/tex], so [tex]\(\sec(x) = 0\)[/tex] is equivalent to:
[tex]\[ \frac{1}{\cos(x)} = 0 \][/tex]
There is no [tex]\(x\)[/tex] for which [tex]\(\cos(x) = \infty\)[/tex]. Hence, [tex]\(\sec(x) = 0\)[/tex] has no solutions.
#### Part B: When [tex]\(\cos(3x) = 0\)[/tex]
[tex]\[ \cos(3x) = 0 \][/tex]
The cosine function is zero at odd multiples of [tex]\(\frac{\pi}{2}\)[/tex]:
[tex]\[ 3x = \left(n + \frac{1}{2}\right)\pi, \quad n \in \mathbb{Z} \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{\left(n + \frac{1}{2}\right)\pi}{3} \][/tex]
### Step 4: Finding Solutions Within the Interval
We need to find the values of [tex]\(x\)[/tex] within the interval [tex]\(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)[/tex].
#### Case [tex]\(n = -1\)[/tex]:
[tex]\[ x = \frac{\left(-1 + \frac{1}{2}\right)\pi}{3} = \frac{-\frac{1}{2}\pi}{3} = -\frac{\pi}{6} \][/tex]
This value is within the interval.
#### Case [tex]\(n = 0\)[/tex]:
[tex]\[ x = \frac{\left(0 + \frac{1}{2}\right)\pi}{3} = \frac{\frac{1}{2}\pi}{3} = \frac{\pi}{6} \][/tex]
This value is within the interval.
### Step 5: Verifying No Other Solutions in the Interval
Let's check if there could be any more values:
- [tex]\(n = -2\)[/tex] gives [tex]\(x = \frac{-3\pi}{6} = -\frac{\pi}{2}\)[/tex] which is an endpoint, not within the open interval.
- [tex]\(n = 1\)[/tex] gives [tex]\(x = \frac{3\pi}{6} = \frac{\pi}{2}\)[/tex] which is an endpoint, not within the open interval.
Hence, the only solutions within the interval [tex]\(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)[/tex] are:
[tex]\[ x = -\frac{\pi}{6} \quad \text{and} \quad x = \frac{\pi}{6} \][/tex]
### Final Answer
The solutions to the equation [tex]\(\sec(x) \cos(3x) = 0\)[/tex] within the interval [tex]\(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)[/tex] are:
[tex]\[ \boxed{-\frac{\pi}{6}, \frac{\pi}{6}} \][/tex]
### Step 1: Understanding the Functions Involved
The given equation involves the secant function [tex]\(\sec(x)\)[/tex] and the cosine function [tex]\(\cos(3x)\)[/tex]. Recall that:
- [tex]\(\sec(x) = \frac{1}{\cos(x)}\)[/tex]
- [tex]\(\cos(kx)\)[/tex] is the cosine function with argument [tex]\(kx\)[/tex], where [tex]\(k\)[/tex] is a constant.
### Step 2: Setting Up the Equation
The equation is:
[tex]\[ \sec(x) \cos(3x) = 0 \][/tex]
For the product of two functions to be zero, at least one of the functions must be zero.
### Step 3: Solving Each Part
Let's solve for each part separately.
#### Part A: When [tex]\(\sec(x) = 0\)[/tex]
[tex]\(\sec(x)\)[/tex] is the reciprocal of [tex]\(\cos(x)\)[/tex], so [tex]\(\sec(x) = 0\)[/tex] is equivalent to:
[tex]\[ \frac{1}{\cos(x)} = 0 \][/tex]
There is no [tex]\(x\)[/tex] for which [tex]\(\cos(x) = \infty\)[/tex]. Hence, [tex]\(\sec(x) = 0\)[/tex] has no solutions.
#### Part B: When [tex]\(\cos(3x) = 0\)[/tex]
[tex]\[ \cos(3x) = 0 \][/tex]
The cosine function is zero at odd multiples of [tex]\(\frac{\pi}{2}\)[/tex]:
[tex]\[ 3x = \left(n + \frac{1}{2}\right)\pi, \quad n \in \mathbb{Z} \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{\left(n + \frac{1}{2}\right)\pi}{3} \][/tex]
### Step 4: Finding Solutions Within the Interval
We need to find the values of [tex]\(x\)[/tex] within the interval [tex]\(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)[/tex].
#### Case [tex]\(n = -1\)[/tex]:
[tex]\[ x = \frac{\left(-1 + \frac{1}{2}\right)\pi}{3} = \frac{-\frac{1}{2}\pi}{3} = -\frac{\pi}{6} \][/tex]
This value is within the interval.
#### Case [tex]\(n = 0\)[/tex]:
[tex]\[ x = \frac{\left(0 + \frac{1}{2}\right)\pi}{3} = \frac{\frac{1}{2}\pi}{3} = \frac{\pi}{6} \][/tex]
This value is within the interval.
### Step 5: Verifying No Other Solutions in the Interval
Let's check if there could be any more values:
- [tex]\(n = -2\)[/tex] gives [tex]\(x = \frac{-3\pi}{6} = -\frac{\pi}{2}\)[/tex] which is an endpoint, not within the open interval.
- [tex]\(n = 1\)[/tex] gives [tex]\(x = \frac{3\pi}{6} = \frac{\pi}{2}\)[/tex] which is an endpoint, not within the open interval.
Hence, the only solutions within the interval [tex]\(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)[/tex] are:
[tex]\[ x = -\frac{\pi}{6} \quad \text{and} \quad x = \frac{\pi}{6} \][/tex]
### Final Answer
The solutions to the equation [tex]\(\sec(x) \cos(3x) = 0\)[/tex] within the interval [tex]\(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)[/tex] are:
[tex]\[ \boxed{-\frac{\pi}{6}, \frac{\pi}{6}} \][/tex]
