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I have this equation : $\tan 2x = 3\cot x$

By rearranging I am getting the solutions: $37.8$, $142$, $218$ and $322$.

However the mark scheme also has $90$, $270$.

Hence I am wondering where I am getting rid of solutions. Here is my working:

$$\frac{2\tan x}{1-\tan^2x} = 3\cot x$$ $$\frac{2\tan^2x}{1 - \tan^2x} = 3$$

and then by rearranging:

$$\tan^2x = \frac{3}{5}$$

$$\tan x = \sqrt{\frac{3}{5}} , -\sqrt{\frac{3}{5}}$$

and then solve using the general solutions of trig equations formula $x = 180n + a$

The mark scheme converts the equation so that $... = 0$ which allows you to make the assumption that the denominator must equal infinity hence giving the other two solutions, however I would guess that it is my multiplication of $\tan x$ which caused the issue here.

UPDATE: This is the solution that is in the mark scheme:

$$\frac{2\tan x}{1-tan^2x} = \frac{3}{\tan x}$$ $$\frac{2\tan^2x - 3 + 3\tan^2x}{(1-\tan^2x)\tan x} = 0$$ $$5\tan^2x - 3 = 0$$ or denominator $= \infty$

$x = 37.8, 218, 142, 322, 90, 270$

Cjen1
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    Your working is fine. I don't get how you can have solutions like 90 and 270. They must be wrong. You are correct. – SchrodingersCat Jun 01 '16 at 11:42
  • @SchrodingersCat I've added the mark scheme solution and I agree with their solution. I just can't work out where the solutions are going in mine. – Cjen1 Jun 01 '16 at 11:47
  • You should verify what happens when dividing by $\cot x$. – N74 Jun 01 '16 at 11:49
  • See, if denominator = $\infty$, then either $\tan x=1$ or $\tan x=0$. Now if $\tan x=1$ then your given equation yields $\infty=3$ and if $\tan x=0$ then your given equation yields $0=\infty$. Do you get it now? – SchrodingersCat Jun 01 '16 at 11:50
  • To obtain $\sin x$, $\cos x$, $\tan x$, $\csc x$, $\sec x$, and $\cot x$, type \sin x, \cos x, \tan x, \csc x, \sec x, and \cot x, respectively, when you are in math mode. – N. F. Taussig Jun 01 '16 at 12:26

1 Answers1

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Hint:

the formula $$ \tan(2x)=\frac{2\tan x}{1-\tan^2x} $$ is not valid for $x=90°+k180°$ and for $\tan x=\pm 1$ so you have to test these values in the given equation.

Emilio Novati
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