Question 558191
I presume you mean *[tex \LARGE \cos 2x + \cos x + 1 = 0] instead of *[tex \LARGE \cos^2 x + \cos x + 1 = 0].


Using trig identities, we have


*[tex \LARGE (\cos^2 x - \sin^2 x) + \cos x = -1]


Regroup this way:


*[tex \LARGE \cos^2 x + \cos x = -1 + \sin^2 x]


Now sin^2 - 1 is equal to -cos^2 x, so we have


*[tex \LARGE \cos^2 x + \cos x = -\cos^2 x]


If cos x = 0 then x = pi/2, 3pi/2, 5pi/2, etc. Otherwise, we may divide by cos x to obtain


*[tex \LARGE \cos x + 1 = -\cos x \Rightarrow \cos x = -\frac{1}{2}]


We obtain x = 2pi/3, 4pi/3, 8pi/3, 10pi/3, ... or in general


*[tex \LARGE x = \frac{2\pi}{3} + 2k\pi] or *[tex \LARGE x = \frac{4\pi}{3} + 2k\pi] or *[tex \LARGE x = \frac{\pi}{2} + k\pi] where k is any integer.