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 $x_2=(3pi)/4$

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