2

How would I solve this trigonometric equation? $$4\cos2x-\sqrt3\cot2x=1$$

I got to this stage, is it a dead end? $$4\sin2x-\sqrt3 = \tan2x$$

LiziPizi
  • 2,855

3 Answers3

2

$\displaystyle4\cos2x-\sqrt3\cot2x=1\iff2\cos2x\sin2x=\sin\dfrac\pi3\cos2x+\cos\dfrac\pi3\sin2x$

$\displaystyle\implies\sin4x=\sin\left(\dfrac\pi3+2x\right)$

$\displaystyle\implies4x=n\pi+(-1)^n\left(\dfrac\pi3+2x\right)$ where $n$ is any integer

See also : Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $

Community
  • 1
lab bhattacharjee
  • 274,582
  • I think there is a $2$ missing in your second form (it should be $2\cos 2x\sin 2x$ in the left-hand side, right? But, your solution is much better than mine!! This should be accepted as the correct solution! – mickep Sep 13 '15 at 12:50
  • @mickep, Thanks for your feedback. Rectified. – lab bhattacharjee Sep 13 '15 at 13:21
1

I don't see an easy way out here. Writing $$ \cos 2x=2\cos^2x-1=\frac{2}{1+\tan^2x}-1 $$ and $$ \cot 2x=\frac{1}{2}\Bigl(\frac{1}{\tan x}-\tan x\bigr) $$ we end up with the equation (where $u=\tan x$) $$ \sqrt{3}u^4-10u^3+6u-\sqrt{3}=0. $$ One solution is $u=1/\sqrt{3}$ which will give you nice solutions for $x$ in the end. The other three solutions to the polynomial equation are real as well, but not as nice. I write them down for you, and you can consider yourself how to continue, $$ \begin{aligned} u_1&=\sqrt{3}-2\sqrt{3}\cos(\pi/9)+2\sin(\pi/9),\\ u_2&=\sqrt{3}+2\sqrt{3}\cos(\pi/9)+2\sin(\pi/9),\\ u_3&=\sqrt{3}-4\sin(\pi/9). \end{aligned} $$

mickep
  • 19,962
0

Hint:

$\cos 2x=\frac{1-tan^2x}{1+tan^2x} $

and, $\cot 2x=\frac{1-tan^2x}{2\tan x}$

now let $1-tan^2x=a$

then

$1+tan^2x=2-a$ and

$2\tan x = 2\sqrt{1-a}$ now you have a quadratic in a.

Display name
  • 1,033