How do you solve the following equation #(cos(t))^2= 3/4# in the interval [0, 2pi]?

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Answer 1 · 2016-09-10T15:50:07

#t = (pi) / (6), (5 pi) / (6), (7 pi) / (6), (11 pi) / (6)#

We have: #(cos(t))^(2) = (3) / (4)#; #[0, 2 pi]#

First, let's take the square root of both sides:

#=> cos(t) = pm sqrt((3) / (4))#

#=> cos(t) = pm (sqrt(3)) / (2)#

Then, the let's determine the basic acute angle:

#=> theta = arccos(sqrt(3) / (2))#

#=> theta = (pi) / (6)#

The interval is given as #[0, 2 pi]#, so the values for #t# will lie in all four quadrants:

#=> t = (pi) / (6), (pi - (pi) / (6)), (pi + (pi) / (6)), (2 pi - (pi) / (6))#

#=> t = (pi) / (6), (5 pi) / (6), (7 pi) / (6), (11 pi) / (6)#