Question

Find exact solutions over the indicated intervals. Written such as xx degrees + 360 deg K. Where k is any integer.

2 cosx + sqrt(3)=0

Answer 1

What part of this is giving you trouble?

Answer 2

Honestly where to even begin.

Answer 3

OK, you have probably seen the unit circle with lots of values on it. Where to start is to get it so cosx= some value you know.

Answer 4

Cos= -sqrt(3)/2?

Answer 5

Exactly! Now, where on the unit circle is it that?

Answer 6

Well I know the positive of that is 30 so I'm not sure if its just -30 or what

Answer 7

It is not just one point. there are two, and generally you use \(0 \le x< 360\)

Answer 8

Once you have the points, you make a formula that would solve for all points.

Answer 9

Did you find the places with that being \(-\frac{\sqrt{3}}{2}\)?

Answer 10

No, I don't know what you mean by that.

Answer 11

In your class, have you looked at sine and cosine as x and y values on the unit circle?

Answer 12

Yes, I have a whole trig table.

Answer 13

Good. Well, then lets talk X and Y for a minute. Forget the trig for just a minute. If I have any four points, \((x,y)\), \((-x,y)\), \((x,-y)\), and \((-x,-y)\) one will be in each quadrant of the Cartesian plane, right? However, notice that x is negative in two quads.

Answer 14

Ok

Answer 15

Now, lets bring back the trig. \(\cos\theta\) is negative anywhere x would be. So the entire left half of the unit circle has negative cosine values. That means there are two points, one in quad II and one in quad III, where \(\cos\theta\) is any particular negative output value.

Answer 16

Ok so 30 and 330?

Answer 17

That would be positive. So what are the same basic positions, but on the negative side of things?