anntasia1987
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# solve cos x + rad 2 = -cos x for over the interval [0, 2pi ]

$cos x + \sqrt{2} = -cosx \\ \\ 2 cos x + \sqrt{2} = 0 \\ \\ cos x = -\frac{\sqrt{2}}{2}$ 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: $x = \frac{3\pi}{4}, \frac{5\pi}{4}$