Question

what is the general solution to sin(x)tan(x)-sin(x)+tan(x)-1=0

Answer 1

Oh this is an interesting problem.
It's really an exercise in factoring.
So this might seem a little tricky at first, see if you can follow along.

\[\large \color{cornflowerblue}{\sin x \tan x} - \sin x + \color{cornflowerblue}{\tan x}-1=0\]We'll start by factoring out a \(\tan x\) from the blue terms.\[\large \tan x(\color{cornflowerblue}{\sin x+1})\color{orangered}{-\sin x-1}=0\]
From here we'll factor out a \(-1\) from each of the orange terms.
\[\large \tan x(\sin x+1)-(\color{orangered}{\sin x+1})=0\]

From here, see how both terms have a \(\sin x+1\)?
Let's factor that chunk out of both terms.\[\large (\sin x+1)(\tan x-1)=0\]

Follow me up to that point?
I know that last step can be a bit tricky.
Let me know if you're confused, we can go further if you understand up to this point.

Answer 2

go ahead

Answer 3

@zepdrix go ahead

Answer 4

So from here we can apply the Zero Factor Property.
We'll set each factor equal to zero separately and solve for x.\[\large \sin x+1=0 \qquad \rightarrow \qquad \sin x=-1\]We need to think back to our unit circle.
What special angle, when we take the sine of it, gives us -1?

I think it's \(\dfrac{3\pi}{2}\), yes?

So we found a solution for x.
\[\large x=\dfrac{3\pi}{2}\]The only problem is, they want the general solution, which includes EVERY TIME \(\sin x=-1\).

So if we spun around the circle a full time, and landed on that same spot on the unit circle, that would also be a solution right? It would still give us -1.

So we can make full revolutions and still get a valid solution.
The way we write that is, we can add or subtract any integer multiple of 2pi.

\[\large x=\dfrac{3\pi}{2} \pm 2k\pi\]

Answer 5

That gives us ONE of the sets of solutions.
We need to get the other from the other factor. The one with the tangent.

We will do the same type of thing.
Set it equal to 0, remember back to your unit circle to recall your reference angles.

Answer 6

what was k agian?

Answer 7

K represents the NUMBER OF TIMES we spin around the circle.
It can be any positive integer.

If we spin around the circle 2 full times, k=2.
We would travel 4pi around, and then land on the 3pi/2 reference angle once again.

Answer 8

i see

Answer 9

@zepdrix is the tan part [\pi/3+k \pi\]?

Answer 10

Tangent is -1 and 1 at all of the pi/4 angles.
So we're concerned with \(\tan x=1\) right?

So if we think back to our special angles, I believe that will occur when \(x=\dfrac{\pi}{4}\) and \(x=\dfrac{5\pi}{4}\).

But we can do away with one of those values since the period of tangent is \(\pi\).
We'll use the \(k\pi\) idea that you were suggesting.

Answer 11

ok, how?

Answer 12

@zepdrix ok how?

Answer 13

Silly computron! C:

You pretty much had the solution already, just with the wrong reference angle.

Answer 14

\[\large x=\frac{\pi}{4}\pm k\pi\]

Answer 15

so i put the to together?

Answer 16

*two

Answer 17

Yes, that sounds right c:
we have 2 sets of solutions for x.

Answer 18

ok, thanks.