Find all solutions of the equation tan^5 x - 9tan x =0.

The answer is Ak\pi where k is any integer,
the constant A= ??.

To find the constant A, let's solve the equation first.

Given equation: tan^5 x - 9tan x = 0

Let's factor out the common term "tan x" from both terms:
tan x (tan^4 x - 9) = 0

Now we have two possible cases:

Case 1: tan x = 0
If tan x = 0, then x can be any multiple of π, so A = 0.

Case 2: tan^4 x - 9 = 0
To solve this quadratic equation, let y = tan^2 x:
y^2 - 9 = 0
(y - 3)(y + 3) = 0

Setting each factor to zero, we have:
y - 3 = 0 or y + 3 = 0

Case 2.1: y - 3 = 0
If y - 3 = 0, then y = 3.

Substituting y back in terms of tan x, we have:
tan^2 x = 3
tan x = ±√3

Using the unit circle or calculator, we find the values of x are π/3 + kπ and 4π/3 + kπ, where k is any integer.

Case 2.2: y + 3 = 0
If y + 3 = 0, then y = -3.

Substituting y back in terms of tan x, we have:
tan^2 x = -3

Since tan x is not defined for values where the denominator is 0, there are no solutions in this case.

Combining the solutions from both cases, we have:
x = π/3 + kπ, 4π/3 + kπ, 0 + kπ

Therefore, the solutions of the equation tan^5 x - 9tan x = 0 can be written as:
x = (π/3 + kπ) or (4π/3 + kπ) or (kπ), where k is any integer.

Therefore, the constant A is 1, since we have a coefficient of 1 in front of π.

To find the values of A, we need to solve the equation tan^5(x) - 9tan(x) = 0.

Step 1: Factor out tan(x) from the equation:
tan(x)(tan^4(x) - 9) = 0

Step 2: Set each factor equal to zero and solve for x:

a) tan(x) = 0
Solution: x = nπ, where n is an integer.

b) tan^4(x) - 9 = 0

We can rewrite tan^4(x) as (tan^2(x))^2:
(tan^2(x))^2 - 9 = 0

Letting u = tan^2(x), we have the quadratic equation:
u^2 - 9 = 0

Step 3: Solve the quadratic equation for u:
u^2 - 9 = 0
(u + 3)(u - 3) = 0
u = -3 or u = 3

Step 4: Substitute back for u to solve for x:
For u = -3:
tan^2(x) = -3
Since tan^2(x) is not negative, there is no solution for this case.

For u = 3:
tan^2(x) = 3
Taking the square root of both sides:
tan(x) = ±√3

To find the solutions, we can use the inverse tangent function:

x = arctan(√3) + mπ where m is an integer
x = arctan(-√3) + nπ where n is an integer

Therefore, the solutions are:
x = arctan(√3) + mπ, x = arctan(-√3) + nπ, and x = nπ

In order to express the solution in the given form Akπ, we need to determine the constant A. The value of A can be found by evaluating the expression tan^5(x) - 9tan(x) at one of the solutions.

For example, substituting x = arctan(√3) + mπ:
A = tan^5(arctan(√3) + mπ) - 9tan(arctan(√3) + mπ)

Using the properties of the tangent function, we can simplify this expression:
A = (√3)^5 - 9(√3)

Simplifying further:
A = 9√3 - 9√3
A = 0

Therefore, the constant A in the given form Akπ is 0.

tan^5 - 9tan = 0

tan(tan^4-9) = 0
tan(tan^2-3)(tan^2+3) = 0
tan(tan-√3)(tan+√3)(tan^2+9) = 0
so,
tan x = 0 or √3 or -√3

now it should be a cinch