# Extraneous solution?

1. Oct 11, 2009

### Mentallic

I attempted to solve $$sin2x=2sin^2x$$ for $$0\leq x\leq \pi$$
as follows:

$$2sinxcosx=2sin^2x$$
$$2sinx(sinx-cosx)=0$$

Therefore,
$$2sinx=0$$ (1)
$$sinx-cosx=0$$ (2)

(2) -- $$cosxtanx-cosx=0$$
$$cosx(tanx-1)=0$$

Therefore,
$$cosx=0$$ (3)
$$tanx-1=0$$ (4)

Hence my solutions should be solving equations (1), (3) and (4).
i.e. $$sinx=0$$, $$x=0,\pi$$
$$cosx=0$$, $$x=\pi/2$$
$$tanx=1$$, $$x=\pi/4$$

Hence, my solution set is $$x=0,\pi/4,\pi/2,\pi$$

But testing the solutions (which I wouldn't have done so in test conditions) show that $$\pi/2$$ does not satisfy the original equation.
This means $$cosx\neq 0$$, but why?
I can't figure out where this extraneous solution came from, and which step I made was invalid to cause this. e.g. I never squared or multiplied the equation by anything etc.

2. Oct 11, 2009

### tiny-tim

Hi Mentallic!

(have a pi: π )
sinx = cosx tanx is not valid if x = π/2

Instead, use sinx - cosx = a multiple of sin(x - something)

3. Oct 11, 2009

### Mentallic

Hi tiny-tim!

At first I was like... what?
Haha now I get it. Thanks for your pi would you also like an e with that pi? :tongue:

Oh yes, the infamous 0/0=something trick. Foiled once again

I never really appreciated the beauty of (I think that method is called the subsidiary technique something or other..), so I've always used alternative solutions, such as squaring the equation and finally removing the extraneous solutions. But, I think that would have been much easier so you have a point there

Thanks tiny-tim, your help is greatly appreciated.