Possible Duplicate:
How to find $x$ in some trigonometric equations
How to solve these trigonometric equations?
$\tan2x-\sin4x = 0$
Possible Duplicate:
How to find $x$ in some trigonometric equations
How to solve these trigonometric equations?
$\tan2x-\sin4x = 0$
HINT: $\sin4x=2\sin2x\cos2x$, and $\tan2x=\dfrac{\sin2x}{\cos2x}$. Now let $a=\sin2x$ and $b=\cos2x$, write your equation in terms of $a$ and $b$, and see what it tells you about $a$ and $b$.
Since this is a homework, some intermediate steps are omitted and left for you to work out.
The identities you need are$^{\dagger}$ $ \sin(4x) = \color{red}{2} \sin(2x) \cos(2x)\\ \tan(2x) = \frac{\sin(2x)}{\cos(2x)} $ Substitute both in $ \tan(2x)-\sin(4x) = 0$ to get $ (1 - 2 \cos^2(2x))\sin(2x) = 0 $ which you should be able to factor into $3$ cases. Solve each case for $x.$
$^{\dagger}$ Fixed error thanks to Thomas Andrews and David Mitra.