Hi, can someone explain me how to solve 2tan(x)=1-tan^2(x) and in the range. [0,2π)

Answer 1

2tan(x) = 1- tan^2(x)

let y = tan(x)

2y = 1-y^2

y^2 +2y = 1

add 1 to both sides

y^2 +2y + 1 = 1+1

(y+1)^2 = 2

take square root of both sides

y+1 = +/-sqr2

y = -1 +/-sqr2

y = tan(x) = -1 +/-sqr2

x = the angle whose tangent = -1 +/-sqr2

x = tan^-1(-1 +/-sqr2)

x = pi/8, 5pi/8, 9pi/8, 13pi/8

or

x = 22.5, 112.5, 202.5, 292.5 degrees

use a calculator with an inverse trig function