## anonymous one year ago Solve the equation algebraically: tanxsinxcosx - 1 = 0 Now, can someone explain why the answer has no solution? Haalllppppp

1. anonymous

perhaps if we rewrite $\tan(x)\cos(x)\sin(x)$ it will be more clear

2. anonymous

since $\tan(x)=\frac{\sin(x)}{\cos(x)}$ you are looking at $\frac{\sin(x)}{\cos(x)}\sin(x)\cos(x)=\sin^2(x)$

3. anonymous

solving $\sin^2(x)-1=0$ is not hard what do you get?

4. anonymous

So we don't factor it like $(\sin(x)-1) (\sin(x)+1) = 0?$

5. anonymous

you can if you like

6. anonymous

or you can just write $\sin^2(x)=1$so $\sin(x)=1$ or $\sin(x)=-1$

7. anonymous

idk apparently the answer doesn't have a solution, that's where i got confused :/

8. anonymous

we are not done yet!

9. anonymous

LOL

10. anonymous

once we solve $\sin(x)=1$ for $$x$$ we can see why there is no solution did you solve it yet?

11. anonymous

How so? I thought the solution can also be $\pi/2$ or $3\pi/2$

12. anonymous

so that makes it undefined?

13. anonymous

right

14. anonymous

since at those number, cosine is zero

15. anonymous

don't forget you stared with $\tan(x)\sin(x)\cos(x)$ and since cosine is zero at those number, a) tangent is not defined there and b) cosine is zero, so you would actually have $$\frac{1}{0}\times 1\times 0$$ moral of the story you cannot cancel zeros

16. anonymous

ahah, that makes a lot of sense now hehe :) tysm!