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Posted by 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! - , Saturday, September 19, 2009 at 8:37am

    sorry, i found my mistake already.

  • Trigo! - , 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

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