2tan^2x-tanx-1=0
0<x<2pi
(2tanx+1)(tanx-1) = 0
tanx = -1/2 or 1
...
Recall in which quadrants tanx is positive or negative
To solve the equation 2tan^2x - tanx - 1 = 0 over the interval 0 < x < 2π, we can use the quadratic formula. The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac))/(2a)
In this equation, a = 2, b = -1, and c = -1.
Plugging these values into the quadratic formula, we have:
x = (-(-1) ± √((-1)^2 - 4(2)(-1)))/(2(2))
= (1 ± √(1 + 8))/4
= (1 ± √9)/4
Now we can simplify the expression under the square root:
√9 = 3
Therefore, our solutions are:
x = (1 + 3)/4 = 4/4 = 1
x = (1 - 3)/4 = -2/4 = -1/2
Since 0 < x < 2π, the solution -1/2 is not within the given interval. Therefore, the only solution in the interval 0 < x < 2π is x = 1.