4
$\begingroup$

The equation to be solve,

$$\tan 2x -\tan x=2$$

My Try :

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

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

$$\frac{2\tan x-\tan x(1-\tan ^2 x)}{1-\tan^2x} =2$$

$$2-2\tan^2x =2\tan x-\tan x+\tan^3x $$

$$\tan^3x+2\tan^2x+\tan x-2=0$$

$\tan x= t$

$$t^3+2t^2+t-2=0$$

now :?

  • 0
    You can just simply solve the polynomial now, no? – 2017-02-25
  • 0
    @qt. how simply ?..... – 2017-02-25
  • 2
    There are no rational roots, so you would have to use the cubic formula or your calculator. The answer is not going to be too clean... – 2017-02-25
  • 1
    Confirmation of the comment above: $$t = \frac{-2 + \sqrt[3]{28 - 3 \sqrt{87}}+ \sqrt[3]{28 + 3 \sqrt{87}}}3\approx0.69562 $$ – 2017-02-25
  • 0
    @Did The other question would be whether the root is maybe a rational multiple of $\pi\,$. – 2017-02-25
  • 0
    @dxiv Is there a name for that? i.e. is there a test for roots that are rational multiples of $\pi$? – 2017-02-25
  • 1
    @zahbaz Don't know that it has a name, and it's a hard problem in general, see for example [When is ArcTan a rational multiple of pi?](http://mathoverflow.net/questions/36272/when-is-arctan-a-rational-multiple-of-pi). – 2017-02-25

2 Answers 2

2

The rational roots theorem gives $\pm(1,2)$ as candidate roots. None of these actually produce zeros. Your recourse is to either rely on the cubic formula or a numerical method, as qt commented.

If you want to explore a numerical solution by hand, you can hone in on the solution with a bit of analysis.

\begin{align} y&=t^3+2t^2+t-2 \\ \\ y'&=3t^2+4t+1 \\ &= (t+1)(3t+1) \end{align}

The function has critical points at $t=-1,-\frac{1}{3}$. We can determine that both $y(-1)<0$ and $y(-1/3)<0$. Given that this is a cubic polynomial, we conclude that $y<0$ (i.e. no zeros) for $t<0$.

Since $y(0)=-2$ and $y'>0$ for $t>0$, we know that there is a single zero for $t>0$. By inspection, it must lie within $(0,1)$. You could try Newton's method from here, if you want to churn out a numerical solution by hand.

0

Now you have to solve the polynomial. Unfortunately, there are no rational roots, so I suggest you to compute this

https://www.wolframalpha.com/input/?i=t%5E3%2B2t%5E2%2Bt-2%3D0

When you got the value of $t$, work again with tangent functions to find the values of $x$ that satisfy the first equation.