Solve for all real values of x.
3tan^2x = 3^1/2 tan x
I have no idea how to approach this problem.
3^1/2 tan x *(3^1/2*tanx - 1)= 0
tan x = 0 or 1/sqrt3
x = 0, pi, pi/6 and one other tan^-1(1/sqrt3)
Much appreciated guys.
To solve the equation 3tan^2(x) = √3tan(x) for all real values of x, we can use a common strategy called substitution. Let's follow these steps:
Step 1: Let's denote tan(x) as a new variable, say u. Therefore, we have:
3u^2 = √3u
Step 2: Now, we can rewrite the equation in quadratic form:
3u^2 - √3u = 0
Step 3: Next, we can factor out the common term u:
u(3u - √3) = 0
Step 4: To find the solutions, we consider two possibilities:
a) Setting the first factor equal to zero:
u = 0
b) Setting the second factor equal to zero and solving for u:
3u - √3 = 0
3u = √3
u = √3/3
Step 5: Since we substituted tan(x) as u, we need to find the corresponding values of x. We can use the inverse tangent function to do this.
a) For u = 0:
tan(x) = 0
x = nπ, where n is an integer.
b) For u = √3/3:
tan(x) = √3/3
x = π/6 + nπ or 7π/6 + nπ, where n is an integer.
Hence, the solution set for all real values of x in the equation 3tan^2(x) = √3tan(x) is:
x = nπ, π/6 + nπ, or 7π/6 + nπ, where n is an integer.
3tan^2x = 3^1/2 tan x
3tan^2x - √3tan x = 0
tanx(3tanx - √3) = 0
so tanx = 0 or tanx = √3/3
if tanx = 0, then x = 0º, 180º, 360º, ...
if tanx = √3/3 then x = 30º, 210º,390º, ...
The period of the tangent function is 180º, so adding 180 to any of our answers produces more answers
the angles in radians would be
0, pi, 2pi ...
pi/6, 7pi/6, 13pi/6, ..
(adding pi each time)