## Trigonometry (11th Edition) Clone

The solution set to this problem is $$\{\pi\}$$
$$\cos^2x+2\cos x+1=0$$ over interval $[0,2\pi)$ 1) Solve the equation: $$\cos^2x+2\cos x+1=0$$ $$(\cos x+1)^2=0$$ $$\cos x+1=0$$ $$\cos x=-1$$ 2) Apply the inverse function: $\cos x=-1$: $\cos x\lt0$ means that the angle of $x$ lies either in quadrant II or III. In fact, looking at the unit circle, we can easily figure out at only $x=\pi$, $\cos x=-1$ Therefore, $$x=\pi$$ In other words, the solution set to this problem is $$\{\pi\}$$