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

Feb 17, 2016

${x}_{1} = \frac{\pi}{4}$ and ${x}_{2} = \frac{3 \pi}{4}$

#### Explanation:

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

${\cos}^{2} \left(x\right) = \frac{1}{2}$ then square root: $\cos \left(x\right) = \frac{1}{\sqrt{2}}$.

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

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

$\frac{1}{\sqrt{2}}$ is the exact value for $\cos \left(\frac{\pi}{4}\right)$
So we know at least ${x}_{1} = {\cos}^{-} 1 \left(\frac{1}{\sqrt{2}}\right) \to {x}_{1} = \frac{\pi}{4}$
From the symmetry of the graph the second value can be obtained by ${x}_{2} = 2 \pi - {x}_{1} = 2 \pi - \frac{\pi}{4} = \frac{3 \pi}{4}$
Thus, within the region, ${x}_{1} = \frac{\pi}{4}$ and ${x}_{2} = \frac{3 \pi}{4}$