# Trig Equation #2

1. Nov 24, 2008

### Draggu

1. The problem statement, all variables and given/known data
Solve for x on the interval 0<= theta <= 2pi. Give exact solutions if possible. Otherwise round to the nearest hundredth of a radian.

2. Relevant equations
tan^2x-4tanx = 0

3. The attempt at a solution

Tried factoring it to (tanx-4)(tanx) = 0

tanx=4 or 0 I guess? Not sure where to go from there

or instead of factoring I could immediately make it sin/cos, which gives

(sin^2x - 4sincos) / (cos^2x) = 0

Last edited: Nov 24, 2008
2. Nov 24, 2008

### gabbagabbahey

I would plot a graph of tan(x) for $x \in [0,\pi]$....at what points does tanx=4?...At what points does tanx=0?...Those will be your solutions.

3. Nov 24, 2008

### Дьявол

"Tried factoring it to (tanx-4)(tanx) = 0"

@Draggu now it is good. So as you said there are two solutions: tanx=4 and tanx=0.

The solutions for x are:

x1=[arctan(4)+kπ]

x2=[arctan(0)+kπ]

Regards.

4. Nov 24, 2008

### Draggu

We have not learned that yet in class... so doing that wouldn't work. :p

5. Nov 24, 2008

### Draggu

There is no exact radian measure/degree where tanx=4
Tanx=4 approx at 76degrees. God I'm lost.

6. Nov 24, 2008

### Draggu

Well I got: (19pi/45), (64pi/45), pi, 2pi as the answers

7. Nov 25, 2008

### Дьявол

You know how to draw graphic of tan function, and do not know how to find arctan(x)?

That's strange.

Anyway, you got the solutions correctly, since the task is to find the solutions in the interval [0,2п]. In the future, you will learn about periods.So if you want to be more correctly you should write 19п/45 + kп and п + kп, where k Є Z

8. Nov 25, 2008

### Chaos2009

Well, for tan(x) = 0, you don't need to know tan(x). $$tan(x) = \frac{sin(x)}{cos(x)}$$ so tan(x) = 0 when sin(x) = 0.