How to solve these trigonometric equations?

$$\tan2x-\sin4x = 0$$


$$\tan2x = \sin x$$

I can't do this, please help me! I did this:

\begin{align} \tan2x &= 2\sin x\\ \\ \frac{\sin2x}{\cos2x} &= \tan x \end{align}

  • $\begingroup$ $0, (1/2)\pi, (1/8)\pi, (3/8)\pi, -(1/8)\pi, -(3/8)\pi;0, \pi, (2/3)\pi, -(2/3)\pi$. $\endgroup$ – i. m. soloveichik Aug 17 '12 at 13:17
  • $\begingroup$ how to do that bro $\endgroup$ – dramasea Aug 17 '12 at 13:20
  • 1
    $\begingroup$ $\sin {2x} = \cfrac {2\tan x} {1+\tan^2 x}$ $\endgroup$ – Mark Bennet Aug 17 '12 at 13:21
  • $\begingroup$ @MarkBennet That is $\tan 2x$, no? $\endgroup$ – process91 Aug 17 '12 at 14:03
  • $\begingroup$ @MichaelBoratko I believe he is using the Weierstrass substitution. $\endgroup$ – preferred_anon Aug 17 '12 at 14:21

$\sin {2x} = \cfrac {2\tan x} {1+\tan^2 x}$

Just to be clear on how to use this substitution:

$$\tan {2x} = \sin {4x} = \cfrac {2\tan 2x} {1+\tan^2 2x}$$

Then we have $\tan {2x} = 0$ or $1+\tan^2 {2x} = 2$ and in this case $\tan {2x} = \pm 1$.

... from which point we are reduced to considering simple cases.

  • $\begingroup$ Mark, actually, I'm a little confused. Why is $\tan 2x = 0$? $\endgroup$ – Andrew Aug 17 '12 at 15:32
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    $\begingroup$ @Andrew - If $\tan {2x}=0$ both sides of the equation are zero, and I have a solution to investigate. If not I can divide through by $\tan {2x}$ to see if there are other possible solutions. $\endgroup$ – Mark Bennet Aug 17 '12 at 15:52

Usually, equations with trigonometric functions in them can be solved by substitution using identities. If you know your trigonometric identities really well, then solving these types of equations becomes less difficult. For the first equation, you have

$$ \tan 2x = \sin 4x$$

In this case, you can use the fact that $\tan x = \cfrac{\sin x}{\cos x}$ to substitute for $\tan 2x$. Also, there's another identity which states that $\sin 2x = 2 \sin x \cos x$, so you can substitute using this identity for $\sin 4x$. This will give you

$$ \cfrac {\sin 2x}{\cos 2x} = 2 \sin 2x \cos 2x$$

You should then be able to solve this equation by multiplying through by $\cos 2x$ and using the identity $\cos^2 {x} = 1 - \sin^2{x}$.

As for the equation $\tan 2x = \sin x$, use the identity that Mark Bennet mentioned in his comment, i.e. that $\tan 2x = \cfrac{2 \tan x}{1 - \tan^2 x}$. You should then have

$$ \cfrac{2 \tan x}{1 - \tan^2 x} = \sin x$$

Using the identity $ \tan x = \cfrac{\sin x}{\cos x}$ and the fact that $\cos^2 x = 1 - \sin^2 x$, you can manipulate the above to produce

$$ \cfrac{2 \cfrac{\sin x}{\cos x}}{1 - \cfrac{\sin^2 x}{1 - \sin^2 x}} = \sin x$$

This simplifies to

$$ \sin {\cfrac{x}{2}} (2 \cos^2 x - 2 \cos x - 1) = 0 $$

Then you just have to check to make sure that solutions to the above equation are well-defined in the original equation.

Edit: Mark's answer is more efficient than mine for the second part of the problem, as he doesn't reduce everything to polynomial functions of sin and cos.

  • $\begingroup$ I fixed a mistake in the final equation. It should be correct now. $\endgroup$ – Andrew Aug 17 '12 at 14:52


$=>\frac{\sin2x}{\cos2x}=2\sin2x \cos2x$



$\sin2x=0=>2x=n\pi=>x=\frac{r\pi}{2}$ for some integer r.

$\cos4x=0=>4x=\frac{(2r+1)\pi}{2}=>x=\frac{(2r+1)\pi}{8}$ for some integer r.

Again, $\sin4x=\sin x$,

$4x=n\pi+(-1)^nx$ for some integer n.

If $n$ is even $=2m(say)=>3x=2m\pi=>x=\frac{2m\pi}{3}$

If n is odd$=2m+1(say)=>5x=(2m+1)\pi=>x=\frac{(2m+1)\pi}{5}$

So we need to resolve r, m in integers such that,

$\frac{r\pi}{2} or \frac{(2r+1)\pi}{8}$ equals to $\frac{2m\pi}{3} or \frac{(2m+1)\pi}{5} $

(1)If $\frac{r\pi}{2} = \frac{2m\pi}{3}=>3r=4m=>r|4=>r=4s$ for some integer s.

$=> x= \frac{r\pi}{2} =\frac{4s\pi}{2} =2s\pi $ .

(2)If $\frac{r\pi}{2} = \frac{(2m+1)\pi}{5}=>5r=(2m+1)2=>5r=4m+2=>5(r-2)=4(m-2)$

$=>4|(r-2)=>r=4s+2$ for some integer s.

$=> x= \frac{r\pi}{2} = \frac{(4s+2)\pi}{2}=(2s+1)\pi$

(3)If $\frac{(2r+1)\pi}{8}=\frac{2m\pi}{3}=>3(2r+1)=16m$ which is impossible as the LHS is odd, the RHS is even for integral r,m.

So, there will be no solution in this case.

(4)If $\frac{(2r+1)\pi}{8}=\frac{(2m+1)\pi}{5}=>5(2r+1)=8(2m+1)$ which is again unsolvable in integers.

So, combining all the 4 cases, $x = t\pi$ where t is any integer(as t is even in case(1), odd in (2)).


$$\tan(2x)-\sin(4x)=0 \implies \tan(2x)=\sin(4x) \implies \frac{\sin(2x)}{\cos(2x)}=\sin(4x)$$ By the double angle formula, $\sin(4x)=2\sin(2x)\cos(2x)$, so assuming $\sin(2x)\ne 0$, $$\frac{1}{\cos(2x)}=2\cos(2x) \implies 2\cos^2(2x)=1 \implies \cos^2(2x)=\frac{1}{2} \implies \cos(2x)=\frac{1}{\sqrt{2}}$$ But we must be careful. When we assumed $\sin(2x)\ne0$, we were implying that $x\ne0,\frac{\pi}{2},\pi$. But $all$ of these are solutions to the equation! So we must include them, too.

If we assume our solutions are in the interval $x \in[0,\pi]$, we have $2x=\frac{\pi}{4},\frac{7\pi}{4}$, so $x=0,\frac{\pi}{8},\frac{\pi}{2},\frac{7\pi}{8},\pi$.

Assuming you understand that, your other equation should be similar. I will leave that as an exercise for you.


Equate tan2x then you will get sinx=sin4x, Now x=n(pi)+((-1)^n)*4x where n is a natural no. For any sinx=siny we have x=n(pi) + ((-1)^n)*y. you can prove this yourself very easily

  • $\begingroup$ What does that mean? $\endgroup$ – SN77 Aug 17 '12 at 13:48
  • $\begingroup$ i don;t know what u mean, complete solution pls $\endgroup$ – dramasea Aug 17 '12 at 13:50
  • $\begingroup$ @dramasea: You're not supposed to give complete solutions to homework questions. It's deemed counterproductive meta.math.stackexchange.com/questions/415/…. $\endgroup$ – Andrew Aug 17 '12 at 15:52

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