Answer :
To find the total number of solutions for the equation [tex]\(\tan(\tan(x)) = \cot(x)\)[/tex] in the interval [tex]\( (0, 2\pi) \)[/tex], let's follow a systematic approach.
### Step-by-Step Solution:
1. Transform the equation:
The given equation is:
[tex]\[ \tan(\tan(x)) = \cot(x) \][/tex]
Recall that [tex]\(\cot(x) = \frac{1}{\tan(x)}\)[/tex]. Substituting this into the equation, we get:
[tex]\[ \tan(\tan(x)) = \frac{1}{\tan(x)} \][/tex]
2. Introduce a substitution:
Let [tex]\( y = \tan(x) \)[/tex]. So, the equation becomes:
[tex]\[ \tan(y) = \frac{1}{y} \][/tex]
3. Solve the transformed equation:
We need to solve:
[tex]\[ \tan(y) = \frac{1}{y} \][/tex]
4. Analyze the solutions for [tex]\( y \)[/tex]:
To solve the equation [tex]\( \tan(y) = \frac{1}{y} \)[/tex], we need to look at the behavior of the functions [tex]\(\tan(y)\)[/tex] and [tex]\(\frac{1}{y}\)[/tex] graphically or analytically.
- [tex]\(\tan(y)\)[/tex] has vertical asymptotes at [tex]\( y = \frac{\pi}{2} + k\pi \)[/tex] for integer [tex]\( k \)[/tex] and is periodic with period [tex]\(\pi\)[/tex].
- [tex]\(\frac{1}{y}\)[/tex] has a vertical asymptote at [tex]\( y = 0 \)[/tex] and approaches [tex]\(\pm \infty\)[/tex] as [tex]\( y \)[/tex] approaches 0 from the positive and negative sides.
On each interval between the vertical asymptotes of the tangent function, [tex]\(\tan(y)\)[/tex] and [tex]\(\frac{1}{y}\)[/tex] intersect once because [tex]\(\frac{1}{y}\)[/tex] is a smooth curve that passes through all four quadrants, whereas [tex]\(\tan(y)\)[/tex] increases monotonically from [tex]\(-\infty\)[/tex] to [tex]\(\infty\)[/tex] on each interval [tex]\(-\frac{\pi}{2} < y < \frac{\pi}{2}\)[/tex].
5. Number of intervals for [tex]\( y \)[/tex]:
In the interval [tex]\( (0, 2\pi) \)[/tex], the possible intervals for [tex]\( y \)[/tex] (i.e., where [tex]\( \tan(x) \)[/tex] must lie) range over two full periods of [tex]\(\pi\)[/tex].
- [tex]\( x \in (0, \pi/2) \implies y \in (-\infty, \infty) \)[/tex] through one period
- [tex]\( x \in (\pi/2, \pi) \implies y \in (-\infty, \infty) \)[/tex]
- [tex]\( x \in (\pi, 3\pi/2) \implies y \in (-\infty, \infty) \)[/tex]
- [tex]\( x \in (3\pi/2, 2\pi) \implies y \in (-\infty, \infty) \)[/tex]
These intervals collectively cover four periods of the tangent function, implying four solutions per period.
6. Count total solutions:
Each period of [tex]\(\pi\)[/tex] in the [tex]\( y \)[/tex]-domain yields one solution due to the properties of the tangent and reciprocal functions.
Since the domain [tex]\( (0, 2\pi) \)[/tex] of [tex]\( x \)[/tex] covers two periods of the [tex]\(\tan(x)\)[/tex], this gives us:
- [tex]\( 2 \text{ solutions} \times 2 \text{ periods} = 4 \text{ total solutions} \)[/tex]
### Conclusion
The total number of solutions to the equation [tex]\(\tan(\tan(x)) = \cot(x)\)[/tex] in the interval [tex]\( (0, 2\pi) \)[/tex] is:
[tex]\[ \boxed{4} \][/tex]
### Step-by-Step Solution:
1. Transform the equation:
The given equation is:
[tex]\[ \tan(\tan(x)) = \cot(x) \][/tex]
Recall that [tex]\(\cot(x) = \frac{1}{\tan(x)}\)[/tex]. Substituting this into the equation, we get:
[tex]\[ \tan(\tan(x)) = \frac{1}{\tan(x)} \][/tex]
2. Introduce a substitution:
Let [tex]\( y = \tan(x) \)[/tex]. So, the equation becomes:
[tex]\[ \tan(y) = \frac{1}{y} \][/tex]
3. Solve the transformed equation:
We need to solve:
[tex]\[ \tan(y) = \frac{1}{y} \][/tex]
4. Analyze the solutions for [tex]\( y \)[/tex]:
To solve the equation [tex]\( \tan(y) = \frac{1}{y} \)[/tex], we need to look at the behavior of the functions [tex]\(\tan(y)\)[/tex] and [tex]\(\frac{1}{y}\)[/tex] graphically or analytically.
- [tex]\(\tan(y)\)[/tex] has vertical asymptotes at [tex]\( y = \frac{\pi}{2} + k\pi \)[/tex] for integer [tex]\( k \)[/tex] and is periodic with period [tex]\(\pi\)[/tex].
- [tex]\(\frac{1}{y}\)[/tex] has a vertical asymptote at [tex]\( y = 0 \)[/tex] and approaches [tex]\(\pm \infty\)[/tex] as [tex]\( y \)[/tex] approaches 0 from the positive and negative sides.
On each interval between the vertical asymptotes of the tangent function, [tex]\(\tan(y)\)[/tex] and [tex]\(\frac{1}{y}\)[/tex] intersect once because [tex]\(\frac{1}{y}\)[/tex] is a smooth curve that passes through all four quadrants, whereas [tex]\(\tan(y)\)[/tex] increases monotonically from [tex]\(-\infty\)[/tex] to [tex]\(\infty\)[/tex] on each interval [tex]\(-\frac{\pi}{2} < y < \frac{\pi}{2}\)[/tex].
5. Number of intervals for [tex]\( y \)[/tex]:
In the interval [tex]\( (0, 2\pi) \)[/tex], the possible intervals for [tex]\( y \)[/tex] (i.e., where [tex]\( \tan(x) \)[/tex] must lie) range over two full periods of [tex]\(\pi\)[/tex].
- [tex]\( x \in (0, \pi/2) \implies y \in (-\infty, \infty) \)[/tex] through one period
- [tex]\( x \in (\pi/2, \pi) \implies y \in (-\infty, \infty) \)[/tex]
- [tex]\( x \in (\pi, 3\pi/2) \implies y \in (-\infty, \infty) \)[/tex]
- [tex]\( x \in (3\pi/2, 2\pi) \implies y \in (-\infty, \infty) \)[/tex]
These intervals collectively cover four periods of the tangent function, implying four solutions per period.
6. Count total solutions:
Each period of [tex]\(\pi\)[/tex] in the [tex]\( y \)[/tex]-domain yields one solution due to the properties of the tangent and reciprocal functions.
Since the domain [tex]\( (0, 2\pi) \)[/tex] of [tex]\( x \)[/tex] covers two periods of the [tex]\(\tan(x)\)[/tex], this gives us:
- [tex]\( 2 \text{ solutions} \times 2 \text{ periods} = 4 \text{ total solutions} \)[/tex]
### Conclusion
The total number of solutions to the equation [tex]\(\tan(\tan(x)) = \cot(x)\)[/tex] in the interval [tex]\( (0, 2\pi) \)[/tex] is:
[tex]\[ \boxed{4} \][/tex]
