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 all solutions of the equation tan^5(x) - 9tan(x) = 0, we can use factoring and trigonometric identities.
Step 1: Factor out the common factor tan(x) from the equation:
tan(x) (tan^4(x) - 9) = 0
Step 2: Set each factor equal to zero:
tan(x) = 0 or tan^4(x) - 9 = 0
Step 3: Solve the first equation, tan(x) = 0:
The solutions for tan(x) = 0 occur when x takes on values of kπ, where k is an integer. You mentioned that the solutions are of the form Akπ, so we can set Akπ = 0 and solve for A.
Akπ = 0
Divide by π: Ak = 0
Since k can be any integer, there are infinite solutions. We can set A = 1 to simplify the form, so the constant A = 1.
Step 4: Solve the second equation, tan^4(x) - 9 = 0:
We can rewrite tan^4(x) as (tan^2(x))^2. Using the trigonometric identity tan^2(x) = sec^2(x) - 1, we have:
(sec^2(x) - 1)^2 - 9 = 0
Simplifying the equation gives us:
(sec^2(x))^2 - 2sec^2(x) + 1 - 9 = 0
(sec^2(x))^2 - 2sec^2(x) - 8 = 0
Substituting u = sec^2(x), we get a quadratic equation:
u^2 - 2u - 8 = 0
Using the quadratic formula, we find the solutions for u:
u = (2 ± √(2^2 - 4(1)(-8))) / (2(1))
Simplifying further, we have:
u = (2 ± √(4 + 32)) / 2
u = (2 ± √36) / 2
u = (2 ± 6) / 2
Solving for u gives us two possible solutions:
u = 4 or u = -2
Now, substitute back u = sec^2(x):
Case 1: u = 4
sec^2(x) = 4
Taking the square root of both sides:
sec(x) = ±2
Using the reciprocal identity for cosine, we have:
cos(x) = ±1/2
The solutions for cos(x) = ±1/2 occur at x = π/3 + 2πk and x = 5π/3 + 2πk, where k is an integer.
Case 2: u = -2
sec^2(x) = -2
Since sec^2(x) cannot be negative, there are no solutions for this case.
Combining the solutions from both cases:
x = π/3 + 2πk, 5π/3 + 2πk, where k is an integer.
Therefore, all solutions of the equation tan^5(x) - 9tan(x) = 0 are given by:
x = π/3 + 2πk, 5π/3 + 2πk, where k is an integer, and A = 1.