Posted by demolitrate on .
solve the following eq. [0<=x<=360]
Well I haven't done trig in a long time but u believe...
You divide the 2 to each side to get
Then do the inverse to both sides to get rid of the cos and end with 3x-72=60 degrees
So x= 44 degrees
Cosine is positive in quad1&4
So 44degrees and 316degrees.
but 3x mean it has turned 360 degree three times, how about the other four ans?
I plugged in 3x-72=___ (set them equal to 420,660,780,1020)
2cos(3x-72) = 1
cos(x-7) = 1/2
we know that cos 60° = 1/2 and cos 300° = 1/2
( (inverse)cos .5) = 60 )
3x-72 = 60 or 3x-72=300
3x = 132 or 3x = 372
x = 44° or x = 124° ---> your basic angles
but the period of cos(3x-72) is 360/3 = 120°
so as long as we don't go beyond your given domain, we can add 120° to any previous answer to get a new answer, so
x = 44+120 = 164
x = 164+120 = 284
x = 284+120 = 404 ---> too big
x = 124+120 = 244
x = 244+120364 -- > too big
so x = 44°, 124°, 164°, 244°, and 284°
just like L.Bianchessi found for you in the above post