Find all the solutions of the equation in the interval {o,2pi)
5sqrt3 tan x+3=8sqrt3 tan x
5√3 tanx + 3 = 8√3 tanx
3√3tanx = 3
√3 tanx = 1
tanx = 1/√3
x = π/6 in quadrant I
or
x = π + π/6 or 7π/6 in quadrant III
x = π/6 , 6π/6
(2sqrt3+8)(5sqrt3-4)
To find the solutions of the equation 5sqrt(3)tan(x) + 3 = 8sqrt(3)tan(x) in the interval [0, 2π), we can follow these steps:
Step 1: Combine like terms by subtracting 5sqrt(3)tan(x) from both sides of the equation:
3 = 3sqrt(3)tan(x)
Step 2: Divide both sides of the equation by 3sqrt(3) to isolate the tangent term:
tan(x) = 1/sqrt(3)
Step 3: Find the reference angle for the tangent of 1/sqrt(3). The reference angle for a tangent value of 1/sqrt(3) is π/6.
Step 4: Determine the solutions in the interval [0, 2π):
Since the tangent function has a period of π, we can add or subtract π from the reference angle to find additional solutions:
Solution 1: x = π/6
Solution 2: x = π/6 + π = 7π/6
Solution 3: x = π/6 + 2π = 13π/6
Therefore, the solutions of the equation 5sqrt(3)tan(x) + 3 = 8sqrt(3)tan(x) in the interval [0, 2π) are x = π/6, 7π/6, and 13π/6.
To find the solutions of the equation 5√3 tan(x) + 3 = 8√3 tan(x) within the interval {0, 2π}, we can use algebraic techniques.
Step 1: Simplify the equation
First, simplify the equation by subtracting 5√3 tan(x) from both sides:
3 = 3√3 tan(x)
Step 2: Isolate the tangent term
Divide both sides of the equation by 3√3:
1/√3 = tan(x)
Step 3: Find the angle whose tangent is 1/√3
The tangent of an angle x equals 1/√3 when x equals either π/6 or 5π/6. These values can be determined by using the unit circle or reference angles.
Step 4: Determine the solutions within the given interval
Since the given interval is {0, 2π}, we need to find the values of x that fall within this range. The solutions are π/6 and 5π/6.
Therefore, the solutions of the equation 5√3 tan(x) + 3 = 8√3 tan(x) within the interval {0, 2π} are x = π/6 and x = 5π/6.