solve the equation on the interval [0, 2pi]:
tan^2 x - 3 tan x + 2 = 0
(tanx - 1)(tanx - 2) = 0
tanx = 1 or tanx = 2
for tanx = 1,
x = π/4 or π+π/4 = 5π/4
for tanx = 2
x = appr 1.107 or π+1.107 = 4.249
To solve the equation tan^2 x - 3 tan x + 2 = 0 on the interval [0, 2pi], we can use the quadratic formula. The quadratic formula states that if we have an equation in the form of ax^2 + bx + c = 0, then the solutions for x can be found using the formula:
x = (-b ± √(b^2 - 4ac))/2a.
In our equation, a = 1, b = -3, and c = 2, since we have tan^2 x - 3 tan x + 2 = 0.
Now, we can substitute these values into the quadratic formula:
x = (-(-3) ± √((-3)^2 - 4(1)(2)))/(2(1))
x = (3 ± √(9 - 8))/2
x = (3 ± √1)/2
x = (3 ± 1)/2
This gives us two possible values for x:
1. x = (3 + 1)/2 = 4/2 = 2
2. x = (3 - 1)/2 = 2/2 = 1
Therefore, the solutions to the equation tan^2 x - 3 tan x + 2 = 0 on the interval [0, 2pi] are x = 1 and x = 2.