3tan^2 (x) + 4tan(x) − 4 = 0

Question

Solve for all solutions in degrees between the range of [0, 360)

Answer 1

3tan²x + 4tanx- 4 = 0. This is quadratic equation for tanx: (tanx + 2)(3tanx - 2) = 0;

tanx = - 2 or tanx = 2/3

tanx = - 2; x = tan^-1(- 2) + 180°n. For n = 1 and n = 2 we have x = 116.565° or x = 296.565°.

tanx = 2/3; x = tan^-1(2/3) + 180°n; for n = 0 and n = 1 we have x = 33.690° or x = 213.690°

Answer 2

Let Z = tan(x)

Then

3Z^2 + 4Z - 4 = 0

(3Z - 2)(Z + 2 ) = 0

3Z-2= 0 or Z +2 = 0

Z = 2/3 or Z=-2

tan x = 2/3

so the first base angle is

x1 = inverse_tangent(2/3)

= 33.69006752597978691352549456166...

The sine is positive in quadrants 1 and 2,

So the solutions from this branch are:

x1 and 180-x1

Next, tan = -2

x2 = inverse_tangent(-2)

x2 = -63.434948822922010648427806279547.. is the base angle

The tangent is negative in quadrants 2 and 4, so the

solutions from this branch are:

x2 and 180 + x2

you can now change the solutions to radians by multiplying

by pi and dividing by 180