How do you solve #Cos^2 x - 1/2 = 0# over the interval 0 to 2pi?

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Answer 1 · 2016-02-17T14:55:44

#x_1=pi/4# and #x_2=(3pi)/4#

First, take the half over to the other side to get:

#cos^2(x) =1/2# then square root: #cos(x)=1/sqrt(2)#.

We now need to find the inverse of this.
If we look at the graph of #cos(x)# over the given region we see:

graph{cos(x) [-0.1,6.15,-1.2,1.2]}

We should expect two answers.

#1/sqrt(2)# is the exact value for #cos(pi/4)#

So we know at least #x_1 = cos^-1(1/sqrt2) ->x_1=pi/4#

From the symmetry of the graph the second value can be obtained by #x_2 =2pi- x_1 = 2pi -pi/4=(3pi)/4#

Thus, within the region, #x_1=pi/4# and #x_2=(3pi)/4#