Trigonometric equation with tangent

[bigplanet401]

Homework Statement

Solve: 2 tan x (tan x - 1) = 3.

Homework Equations

Pythagorean identities?

The Attempt at a Solution

I tried the following:

2 tan^2 x - 2 tan x = 3
2 (sec^2 x - 1) - 2 tan x = 3
2 (1 - cos^2 x) - 2 sin x cos x = 3 cos^2 x (multiplying through by cos^2 x)

...but then I got confused (not wanting to write sin x = sqrt(1 - cos^2 x) and end up with an algebraic mess). Is there an easier way to solve the problem? Thanks!

 

[Dick]

Homework Statement

Solve: 2 tan x (tan x - 1) = 3.

Homework Equations

Pythagorean identities?

The Attempt at a Solution

I tried the following:

2 tan^2 x - 2 tan x = 3
2 (sec^2 x - 1) - 2 tan x = 3
2 (1 - cos^2 x) - 2 sin x cos x = 3 cos^2 x (multiplying through by cos^2 x)

...but then I got confused (not wanting to write sin x = sqrt(1 - cos^2 x) and end up with an algebraic mess). Is there an easier way to solve the problem? Thanks!

Use the quadratic equation. Substitute u=tan(x). Now your first equation is 2u^2-2u=3. Solve that for u. Then find x.

 

[interhacker]

Solve the equation as a quadratic, initially.

$y = \tan x \\ \Rightarrow 2y(y-1) = 3 \\ \Rightarrow 2y^2 - 2y - 3 = 0 \\ \Rightarrow y = \frac {2 \pm \sqrt {28}}{4} \\ \Rightarrow y = \frac {2 \pm 2\sqrt{7}}{4} \\ \Rightarrow y = \frac {1 \pm \sqrt{7}}{2}\\$

Then deal with the trig ratio.

$\tan x = \frac {1 \pm \sqrt{7}}{2} \\ \Rightarrow x = \arctan \frac {1 + \sqrt{7}}{2} \| x = \arctan \frac {1 - \sqrt{7}}{2}\\ Between\, the\, interval:\\ 0 < x < 2\Pi\\ x = 1.07\, rad., 4.20\, rad., 2.45\, rad., 5.59\, rad.$