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trig question....help

can somebody please explain how they would solve this for x aswell please? : cos(3x+40)=0.2
This is what I did, if this is wrong please explain why aswell (thax_) : cos-1(0.2)= 3x+40, so x= 12.83 and you increase the limits. If you sub that into the original equation it comes out as 0.2 but when I put on a cast diagram I cant seem to find any other values for x that give you 0.2 wen u sub it in :s-smilie:

INTERVAL: 0<x<180 so you change them to 0<x<580

Reply 1

MissBrown

To clarify

You have taken the inverse cosine of both sides

Obtained a principle value

Set 3x+40 to that value

Found (by whichever method) the other values in the interval for 3x+40 checking for all that may fall into interval

You have then subbed 40 from each value

Split your answers by 3 and Checked they are all applicable

Correct?

Reply 2

nmudz_009

Original post by MissBrown

To clarify

You have taken the inverse cosine of both sides

Obtained a principle value

Set 3x+40 to that value

Found (by whichever method) the other values in the interval for 3x+40 checking for all that may fall into interval

You have then subbed 40 from each value

Split your answers by 3 and Checked they are all applicable

Correct?

yeah. Just spent another half hour on it an got: x=12.83, x=438.46, x=372.8. Same method, just found the values in a different way. dnt know if its ryt

Reply 3

MissBrown

Original post by nmudz_009

yeah. Just spent another half hour on it an got: x=12.83, x=438.46, x=372.8. Same method, just found the values in a different way. dnt know if its ryt

Hi
Remember your final solutions need to be in the original interval

So you have done:

3x+40 = cos^-1(.02)
3x+40 = 78.46
3x = 38.46
x = 12.8 degrees to 3 sf

and found the first

Now, consider that 3x+40 = 360-principle value
and consider 3x+40 = principle value + 360

See below

http://imageshack.us/photo/my-images/221/trigl.jpg/

Try from that to work out the values you need. All must fall into original interval