[tex]cos x + \sqrt{2} = -cosx \\ \\ 2 cos x + \sqrt{2} = 0 \\ \\ cos x = -\frac{\sqrt{2}}{2} [/tex]
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
[tex]cos (\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} \\ \\ cos (\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}[/tex]
Final Answer:
[tex]x = \frac{3\pi}{4}, \frac{5\pi}{4} [/tex]
