# Sinxtanx - tanx = 0 ; Solve for x?

• Jan 15th 2013, 11:09 PM
Pourdo
Sinxtanx - tanx = 0 ; Solve for x?
Hey everyone!

Sinxtanx - tanx = 0 ; Solve for x? 0 < x < 2pi (and including 0 and 2pi, not sure how to input that)

This is what I have done so far--
sinxtanx - tanx = 0

tanx (sinx - 1) = 0

Set tanx = 0 and sinx - 1 = 0

tanx = 0 when x = 0, pi, 2pi

sinx - 1 = 0
sinx = 1
x = pi/2

so x = 0, pi/2, pi, 2pi

What I am not sure about, is if pi/2 is a solution or not, as I am thinking that tanx is undefined at pi/2.
• Jan 15th 2013, 11:51 PM
abender
Re: Sinxtanx - tanx = 0 ; Solve for x?
$\sin(x)\tan(x)-tan(x)=0 \implies \tan(x)[\sin(x)-1]=0 \implies$ $\frac{\sin(x)}{\cos(x)}\left[\sin(x)-1\right]=0$

You want to find $x$ within your given boundaries where $\sin(x)=0$ and $\sin(x)=1$, AS LONG AS $\cos(x)\neq0$.

Edit: To answer your apparent question, you are right, $x=\tfrac{\pi}{2}$ is NOT a solution, because $\cos(\tfrac{\pi}{2})=0$, which cannot be in a denominator.
• Jan 15th 2013, 11:55 PM
Pourdo
Re: Sinxtanx - tanx = 0 ; Solve for x?
So because cos(x) = 0 at pi/2, it is not a solution?
• Jan 15th 2013, 11:59 PM
abender
Re: Sinxtanx - tanx = 0 ; Solve for x?
Quote:

Originally Posted by Pourdo
So because cos(x) = 0 at pi/2, it is not a solution?

Correct. Recall that $\tan(x)=\frac{\sin(x)}{\cos(x)}$.