2tan^2x-tanx-1=0

0<x<2pi

•MATH - Steve, Thursday, July 18, 2013 at 11:48am
(2tanx+1)(tanx-1) = 0
tanx = -1/2 or 1
...
Recall in which quadrants tanx is positive or negative

....................would x= 5pi/6,11pi/6,pi/4 and 5pi/4???

no, arctan(1/2) = .46

so, x = pi+.46 or 2pi-.46

the pi/4 and 5pi/4 are correct

(arcsin(1/2) = pi/6, not arctan(1/2))

small typo, I am sure Steve meant to type

x = π - .46 or 2π-.46 , since the tan is negative in II and IV

Good catch - what was I thinking?

But then, I'm sure Robin knew that...

in radians what would the .46 be?

That IS in radians.

To solve the equation 2tan^2x - tanx - 1 = 0, you can use the factoring method. Rearrange the equation to get:

(2tanx + 1)(tanx - 1) = 0

Now, set each factor equal to zero:

2tanx + 1 = 0 or tanx - 1 = 0

Solve each equation separately:

For 2tanx + 1 = 0,
2tanx = -1
tanx = -1/2

For tanx - 1 = 0,
tanx = 1

Now, let's verify the possible values of x within the range 0 < x < 2π.

tanx = -1/2 between 0 and 2π:
To find the values of x that satisfy this equation, we need to recall the quadrants in which tanx is positive or negative. tangent is positive in the first and third quadrants. So, we look for angles within these quadrants that have a tangent value of -1/2. Two such angles are 5π/6 and 11π/6.

tanx = 1 between 0 and 2π:
This equation has a solution in the first quadrant where tanx is positive. One such angle is π/4.

So, the possible values of x that satisfy the equation 2tan^2x - tanx - 1 = 0 within the range of 0 < x < 2π are: x = 5π/6, 11π/6, and π/4.