How do you solve $$2 \cos x \sin x - \sin x=0$$ for $$ 0 \le x

Question

How do you solve $$2 \cos x \sin x - \sin x=0$$  for $$ 0 \le x <2\pi$$

I solved and got $$\frac{ \pi }{ 3 }\  and   \frac{ 5\pi }{ 3 }$$

My teacher said I should also include $$0$$ and   $$\pi$$

Can anyone explain to me why?

zehanz · Answer

If you factor out the sinx you get:$$2\cos x \sin x-\sin x=0 \Leftrightarrow \sin x(2\cos x-1)=0$$Now, a product is zero if one or more factors of it are zero, so:$$\sin x=0 \vee 2\cos x -1 =0$$
sin x = 0 has 0, pi and 2pi as solutions.

zehanz · Answer

2pi not included, because x < 2pi

zehanz · Answer

You probably divided by sin x, thereby letting two more solutions magically disappear!

anonymous · Answer

That is exactly what I did.  Thanks.  How would I know not to do that?

zehanz · Answer

Dividing by something unknown is not a good idea: 
sin x can be anything between -1 and 1. 
Mostly it is not equal to zero, but if it is, you can't divide by it!
And when it is 0 ther you have a solution:

Just remember if you've got an equation of the form ab=0: $$a \cdot b=0 \Leftrightarrow a=0 \vee b=0$$

anonymous · Answer

Awesome explanation - if I could give you a hundred medals, I would!

zehanz · Answer

Hear, hear...