How do you solve this trig problem?

Question

2tan^2(x) - 4tan(x) - 1 = 0, 0 ≤ x ≤ 2π

Answer 1

what i would do is use substitution

let tanx = x


2x^2 - 4x - 1 = 0

use quadractic formula
-b +/- sqrt(b^2 - 4ac) / 2

4 +/- sqrt(4^2 - 4(2)(-1)) / (4)

4 +/- sqrt(24) / 4

4 +/- (2x2 x 6)

4 +/- 2 sqrt(6) / 4

2 +/- sqrt(6) / 2


tanx = 2+sqrt(6) / 2
x = 1.148366812 and 4.289959466

tanx = 2 - sqrt(6)/2
x = 2.92052106 and 6.0621137

Answer 2

Let y=tan(x), then solve:

2y^2 - 4y - 1 = 0

y = (4 +/- sqrt(16 + 8))/4
y = 1 +/- sqrt(24)/4
y = 1 +/- sqrt(6)/2

Now that you have two solutions for y, use the first equation (y=tan(x) --> x=tan^-1(y)) to find the values for x which yield each solution:

(a) tan^-1(1 + sqrt(6)/2)
tan^-1(2.2247...)
x = 1.148367 + k*π, where k is an integer.

The solutions between 0 and 2π are 1.148367 and 4.28996 (k=0 and k=1)

(b) tan^-1(1 - sqrt(6)/2)
tan^-1(-0.2247...)
x = -0.2210716 + k*π, where k is an integer.

The solutions between 0 and 2π are 2.9205 and 6.06211 (k=1 and k=2).

Those are the four values of x between 0 and 2π which yield the appropriate two values of y that solve the quadratic.

Answer 3

1. isolate 1
2tan(tan-2)=1
2. solve for each term.
2tan=1, tan=1/2, inverse tangent of 1/2.
tan-2=1, tan=3, inverse tangent of 3.

Answer 4

factor the quadratic and then solve for the roots
ex:

x^2 +2x +1 => (x+1)*(x+1)

Answer 5

Let y = tan(x)
2y^2 - 4y - 1 = 0
Solve for y.
For each y, calculate x = arctan(y) that lies in the given domain.

Answer 6

That's a quadratic

tanx = 1 +/- sqrt(1.5)
= -0.2247, 2.2247

x = 1.1484, 2.9205, 4.2900, 6.0621