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July 29, 2014

July 29, 2014

Posted by **help!** on Saturday, September 19, 2009 at 8:35am.

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 8:37amsorry, i found my mistake already.

- Trigo! -
**Anonymous**, Saturday, September 19, 2009 at 8:43amI 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

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