Posted by help! on Saturday, September 19, 2009 at 8:35am.
Ok, i've tried this sum a million times and i cant get an answer. Somebody please check my workings and tell me what i did wrong! Thanks!
Given 4≤y≤6, find the value of y for which 2cos((2y)/3) + √3 = 0
Here's my workings:
2.666≤(2y/3)≤4
2cos(2y/3)=√3
cos(2y/3)= (√3)/2
ref angle: 2.6179
2y/3= 0.5236, 5.759
and i can't seem to continue, because the answers don't tally with the range i calculated. ???
the correct answer is supposed to be y=5.5
please tell me what i did wrong!

Trigo!  help!, Saturday, September 19, 2009 at 8:37am
sorry, i found my mistake already.

Trigo!  Anonymous, Saturday, September 19, 2009 at 8:43am
I would start this way,
cos (2y/3) = √3/2
so 2y/3 is in quadrants II or III
the reference angle is .5236 radians
so 2y/3 = pi  .5236 OR 2y/3 = pi + .5236
so y = 3.92699 or y = 5.4978
and the value which falls in your given range of y is 5.4978 or appr. 5.5