Solution:
Given:
[tex]\begin{gathered} \cos (\theta)=-\frac{\sqrt[]{2}}{2} \\ \text{where;} \\ 0\le\theta\le\pi \end{gathered}[/tex][tex]\begin{gathered} \cos (\theta)=-\frac{\sqrt[]{2}}{2} \\ \text{This shows the cosine of the angle has a negative value.} \\ Co\sin e\text{ is negative in the second and third quadrants.} \\ \text{Thus,} \\ \cos (\theta)=-\frac{\sqrt[]{2}}{2} \\ \cos \theta=\frac{\sqrt[]{2}}{2} \\ \cos \theta=\frac{1.4142}{2} \\ \cos \theta=0.7071 \\ \theta=\cos ^{-1}(0.7071) \\ \theta=45^0 \end{gathered}[/tex]
In the second quadrant,
[tex]\begin{gathered} \theta=180-\theta \\ \theta=180-45 \\ \theta=135^0 \\ \\ \text{Taking this to radians,} \\ 2\pi\text{ radians=360 degr}ees \\ 2\pi=360^0 \\ x=135^0 \\ \text{Hence, the angle 135 degre}es\text{ in radians is,} \\ x=\frac{2\times\pi\times135}{360} \\ x=\frac{270\pi}{360}radians \\ x=\frac{3\pi}{4}\text{radians} \\ \\ \text{Hence, } \\ \theta=\frac{3\pi}{4}\text{radians} \end{gathered}[/tex]
We do not need the angle in the third quadrant since the range given is between 0 to 180 degrees and the third quadrant exceeds 180 degrees.
Therefore, the value of the angle in radians in simplified rationalized form is;
[tex]\theta=\frac{3\pi}{4}\text{radians}[/tex]
