# Solving

• Dec 15th 2007, 02:40 PM
Hasan1
Solving
$\tan2x = 8 \cos^2x - \cot x [0,\frac{\pi}{2}]$
• Dec 15th 2007, 04:08 PM
Soltras
Arrange it all on one side so it's

$\tan 2x + \cot x - 8 \cos^2 x = 0$.

Now the strategy is to factor the left side to get smaller pieces that are forced to be equal to zero.

Start by recalling $\tan 2x = \frac{\sin 2x}{\cos 2x}$ and $\cot x = \frac{\cos x}{\sin x}$, so multiply through by $\cos 2x \sin x$.

You get:

$\sin 2x \sin x + \cos x \cos 2x - 8 \cos^2 x \cos 2x \sin x = 0$.

Now remember $\sin 2x = 2\sin x \cos x$. So:

$2 \sin^2 x \cos x + \cos x \cos 2x - 8 \cos^2 x \cos 2x \sin x = 0$.

And now we can make our first factor, which is $\cos x$.

$\cos x \left(2 \sin^2 x + \cos 2x - 8 \cos x \cos 2x \sin x \right) = 0$.

No we know $\cos x = 0$ when $x = \frac{\pi}{2}$ (in your range), so that is one solution. Let's forget about that factor now and concentrate on the rest:

$2 \sin^2 x + \cos 2x - 8 \cos x \cos 2x \sin x = 0$.

Recall that $\cos 2x = \cos^2 x - \sin^2 x$, and apply this to the middle term:

$2 \sin^2 x + \cos^2 x - \sin^2 x - 8 \cos x \cos 2x \sin x = 0$

Simplifies to:

$\sin^2 x + \cos^2 x - 8 \cos x \cos 2x \sin x = 0$.

Now, remember that $\sin^2 x + \cos^2 x = 1$, so we now have:

$1 - 8 \cos x \cos 2x \sin x = 0$.

But refer again to the fact that $2 \sin x \cos x = \sin 2x$. so we have:

$1 - 4 \sin 2x \cos 2x = 0$.

Apply the above identity again:

$1 - 2 \sin 4x = 0$.

Now solve for $x$ by just isolating it the old fashioned way. From this, you should get two more values in the interval $0$ to $\frac{\pi}{2}$, making three values total because of the one we found earlier.