Question

Solve cos x + rad 2 = -cos x for over the interval [0, 2pi ]

Answer 1

$cos x + \sqrt{2} = -cosx \\ \\ 2 cos x + \sqrt{2} = 0 \\ \\ cos x = -\frac{\sqrt{2}}{2}$

The cosine function is negative in 2nd and 3rd quadrants, So you know that you will have 2 solutions. One between pi/2 and pi, the other between pi and 3pi/2.

Refer to a unit circle to find that 
$cos (\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} \\ \\ cos (\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}$

Final Answer:
$x = \frac{3\pi}{4}, \frac{5\pi}{4}$