solve the following over the interval [0,2pi)?

Question

2tan x / 3-2tan^2x =1

Answer 1

Interpretation # 1

2/3 tan x - 2 tan² x =1

so 6 tan² x - 2 tanx + 3 = 0

Discriminant < 0 so no real solutions

Interpretation # 2

2 tan (x/3) - 2 tan² x =1

tan 3t = tan (t + 2t)
= (tan t + tan 2t)/(1 - tan t tan 2t)
= (tan t + 2tan t/1 - tan² t)/( 1 - tan t*2tan t/1 - tan² t)
= (tan t (1 - tan² t) + 2tan t)/ (1 - tan² t - 2tan² t)
= (3tan t - tan³ t)/(1 - 3tan² t)

Let tan x/3 = p Then tan x = p(3 - p²)/(1 - 3p²)
So 2 tan (x/3) - 2 tan² x
= 2p - 2 (p(3 - p²)/(1 - 3p²))²

= [2p(1 - 3p²)² - 2p²(3 - p²)²]/(1 - 3p²)² = 1

So 2p[(1 - 3p²)² - p(3 - p²)²] = (1 - 3p²)²

ie (2p - 1)(1 - 3p²)² - 2p²(3 - p²)² = 0

So (2p - 1)(1 - 6p² + 9p^4) - 2p²(9 - 6p² + p^4) = 0

So 18p^5 - 9p^4 - 12p³ + 6p² + 2p - 1 - 2p^6 + 12p^4 - 18p² = 0

ie 2p^6 - 18p^5 - 3p^4 + 12p³ + 12p² - 2p + 1 = 0

So p = tan(x/3) ≈ 1.09073, 9.08455 (by equation solver)

Since 0 ≤ x < 2π
then 0 ≤ x/3 < 2π/3

So x/3 ≈ 0.82877, 1.46116
So x ≈ 2.4863, 4.3835

Answer 2

If 3 - 2 tan² x not = 0, multiplying by it does not ruin the equation. You get
2 tan x = 3 - 2 tan² x, so
2 tan² x + 2 tan x - 3 = 0 which is quadratic.

Never mind, you need parentheses. Is it
2 tan (x/3) - 2 tan² x = 1 or is it
(2 tan x)/3 - 2 tan² x = 1 or is it
(2 tan x) / (3 - 2 tan² x) = 1 ???

Answer 3

2/3 tan x - 2 tan² x =1

so 6 tan² x - 2 tanx + 3 = 0

a=2/3, b=-2, c=3

b^2-4ac = -68

Discriminant b^2-4ac<0 is so no real solutions for that question.

Answer 4

substitute tanx =t, sec^2xdx =dt

so the integral becomes: In{ 2tdt/{(3-t^2)(1+t^2)}

now proceed yourself by solving the denominator and then using partial sums.

Answer 5

need more info like values between, integral, derivative, etc.

Answer 6

-2 ....?