Answer:
[tex]120\º \text{ and } 240\º[/tex]
or in radians:
[tex]$\frac{2\pi }{3} \text{ and } \frac{4\pi }{3}$[/tex]
Step-by-step explanation:
From the way you wrote, you want to solve the equation
[tex]\sqrt {2 \cos(x)+1}=0[/tex] for [tex]0 \leq x < 360\º[/tex], or in radians [tex][0, 2\pi)[/tex]
[tex]\sqrt {2 \cos(x)+1}=0[/tex]
Square both sides
[tex]2 \cos(x) +1= 0[/tex]
[tex]2\cos(x)=-1[/tex]
[tex]$\cos(x)=-\frac{1}{2} $[/tex]
In the Unit Circle, considering one revolution (interval [tex][0, 2\pi)[/tex]),
the values where [tex]$\cos(x)=-\frac{1}{2} $[/tex] are in Quadrant II and III.
Once
[tex]$\cos(x)= \frac{1}{2} \text{ for } x = 60\º \text{ in the Quadrant I}$[/tex]
The values where
[tex]$\cos(x)=-\frac{1}{2} $[/tex] are 120º and 240º.
