Thread: Sinxtanx - tanx = 0 ; Solve for x?

1. 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.

2. Re: Sinxtanx - tanx = 0 ; Solve for x?

$\displaystyle \sin(x)\tan(x)-tan(x)=0 \implies \tan(x)[\sin(x)-1]=0 \implies$$\displaystyle \frac{\sin(x)}{\cos(x)}\left[\sin(x)-1\right]=0$

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

Edit: To answer your apparent question, you are right, $\displaystyle x=\tfrac{\pi}{2}$ is NOT a solution, because $\displaystyle \cos(\tfrac{\pi}{2})=0$, which cannot be in a denominator.

3. Re: Sinxtanx - tanx = 0 ; Solve for x?

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

4. Re: Sinxtanx - tanx = 0 ; Solve for x?

Originally Posted by Pourdo
So because cos(x) = 0 at pi/2, it is not a solution?
Correct. Recall that $\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}$.

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2tanxsinx-tanx=0

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