'Further trigonometric equations' - Need help understanding question procedure

[Harry_the_cat]

I got cos x = -(sqrt2)/8

 

[Inertia_Squared]

ahh yeah I see what I did, I treated 8 as sqrt8 and tried to cancel the sqrt2s, yes that is correct I just got a little ahead of myself ?

 

[Harry_the_cat]

Also look back at your question. Are you working in degrees or radians?

 

[Inertia_Squared]

yeah sorry its a force of habit, I'll be working on it though!

 

[Harry_the_cat]

Ok are you right from here?

 

[Inertia_Squared]

I think so, just to check, given cosx = -sqrt2/8, x ~ 1.75rad, so will x be equal to 1.75rad and 3pi/2 - 1.75rad?

 

[Inertia_Squared]

found the answer for the example question, and it looks to be right, thanks so much for the help and your time! I know I was probably a little frustrating but I really do appreciate the support. Have a good one!

 

[Harry_the_cat]

Not quite. 1.75 rad is correct but the other one isn't.
I prefer to do it this way. Forget the neg sign for the moment and calculate invcos((sqrt2)/8) to get 1.39rad

 

[Inertia_Squared]

ah I see, yeah you're right I read the answer wrong, man I really am more tired than I thought, but that makes sense, thanks again

 

[Harry_the_cat]

Then add/subtract this from 0, pi, or 2pi to place it in the quadrant where cos is neg.
So here pi-1.39 and pi+1.39.

 

[Harry_the_cat]

Are you in Australia? NSW?

 

[Inertia_Squared]

I think that I've finally got it

 

[Inertia_Squared]

yeah, 10:30 rn but I've been pulling 1:00 ams the past 3/4 nights and I get up at 10 to 6 each morning

 

[Harry_the_cat]

Good night! Go to sleep!

 

[Inertia_Squared]

haha will do! goodnight!