Thread: Simple Trigonometric Equations

These are hard for me right now, it will be fine once I understand it but I'm not sure because it's very confusing!

If I had cos 3x = 0.5, would I:
Use cos^-1 0.5 = 60
and because cos is positive in the second quadrant, would I add the 60 to 90 to make 150...
then divide both answers by 3 to get 20 and 50?

Is this right, and would it work for any other including sin?

Also if I had tan x = -root 3, would I solve it like:
tan^-1 root 3 = 60
and because tan is positive in the third quadrant, would I add the 60 again to 180 to make 240
and are my solutions 60 and 240, and nothing else?

Hello db5vry

Originally Posted by db5vry

If I had cos 3x = 0.5, would I:
Use cos^-1 0.5 = 60
and because cos is positive in the second quadrant, would I add the 60 to 90 to make 150...
then divide both answers by 3 to get 20 and 50?

Look at the attached diagram, showing . The other angle in the diagram whose cosine is also is in the fourth quadrant: it is .

If we allow angles of more than , the next angle whose cosine is 0.5 is ; the next is , and so on.

So if , the possible values of are

To find the values of , then, you'll need to divide each of these by to get:



(Cosine is positive in the first and fourth quadrants, of course.)

Grandad

Originally Posted by Grandad

Hello db5vryLook at the attached diagram, showing . The other angle in the diagram whose cosine is also is in the fourth quadrant: it is .

If we allow angles of more than , the next angle whose cosine is 0.5 is ; the next is , and so on.

So if , the possible values of are

To find the values of , then, you'll need to divide each of these by to get:



(Cosine is positive in the first and fourth quadrants, of course.)

Grandad

Thank you for this help!

Of course I remember cosine is positive in the fourth now! It's sine that is positive in the second, don't know why I thought that..
But that was excellent thanks very much!