Missing root of equation $\tan(2x)=2\sin(x)$

I tried to solve the equation $\tan(2x)=2\sin x$ and got the roots $x=n2\pi$ and $x=\pm \frac {2\pi}3 +n2\pi$. It seems that $x=n\pi$ is also a root but for some reason I didn't get that one out of my equation. Could you tell me where I went wrong? $$\tan(2x)=2\sin x$$ $$\frac{(\sin 2x)}{(\cos 2x)}=2\sin x$$ $$\frac{(2\sin x \cos x)}{((\cos x)^2-(\sin x)^2)}=2\sin x$$ $$2\sin x\cos x=2\sin x\cos^2 x-2\sin^3 x$$ $$\sin x\cos x=\sin x(\cos^2 x-\sin^2x)$$ $$\cos x=\cos^2 x-\sin^2 x$$ $$\cos x=\cos^2 x+\cos^2x-1$$ $$2\cos^2x-\cos x-1=0$$

Solving this quadratic equation gives $\cos x=1$ or $\cos x=-\frac 12$ $$\cos x=1=\cos \theta$$ $$x=n2\pi$$ $$\cos x=-\frac 12=\cos(\frac{2\pi}3)$$ $$x=\pm\frac{2\pi}3+n2\pi$$

• What about $\sin(x)=0$ ? – Claude Leibovici Sep 28 '14 at 13:46
• The fifth line looks strange ! – Claude Leibovici Sep 28 '14 at 13:49
• I'm sure you understand that you cannot divide by zero. But also very important: don't divide by anything that even might be zero. . . at least not without considering that as a separate possibility. – David Sep 28 '14 at 13:50

Classic mistake occurs at the line where you go to cancel a $\sin x$ from each side...that only holds if $\sin x$ isn't 0. You also get answers whenever $\sin x=0$, which occurs at increments of $n\pi$

You can't just cancel out $\sin(x)$ in your steps because $\sin(x)$ can also be zero so you have to factor out even $\sin(x)$. $\sin(x)=0$ implies $x=n \pi$ is also the solution.

We have $2\sin x\cos2x=\sin2x=2\sin x\cos x\iff2\sin x(\cos2x-\cos x)=0$

$\sin x=0\implies x=m\pi\ \ \ \ (1)$ where $m$ is any integer

else $\cos2x-\cos x=0\iff\cos2x=\cos x\implies2x=2r\pi\pm x$ where $r$ is any integer

Taking the '+' sign, $x=r\pi$ which is same as $(1)$

Taking the '-' sign, $2x=2r\pi-x\iff x=\dfrac{2r\pi}3$

with $\tan(2x)=\frac{2\cos(x)\sin(x)}{\cos(x)^2-\sin(x)^2}$ we get $2\sin(x)\left(\frac{\cos(x)}{\cos(x)^2-\sin(x)^2}-1\right)=0$ thus we obtain $\sin(x)=0$ or $0=2\cos(x)^2-\cos(x)-1$

Just to expand on all the other correct answers:

If, following Line $\#5$ you had multiplied out the RHS, and then brought everything to the same side, and factored, you would have had a single term (with two complex factors) that equalled zero. Setting each factor in turn equal to zero would of necessity produced all the roots.

None of the above steps have any restrictions. Division by a factor that could be zero produces a branching in the solution...