Why does $\cos(x)(\cos(x) - 1) = 0$ have solutions $x=\pi/2$ and $x=3\pi/2$ (as well as $x=0$)? The original equation is
$$2 - \sin^2(x) = 2 \cos^2 (x/2) \tag{1}$$
I used the Half-angle formula for $\cos$ as well as the Pythagorean identity for sine and cosine to reduce it and ended up with
$$\cos(x)(\cos(x) - 1) = 0 \tag{2}$$
The book has the answers as $x=\pi/2, 3\pi/2, 0$. I understand the $0$, but I thought the other answer would be $\pi$ since on the unit circle $\pi$ has a cosine of $1$?
If it's not mentioned that $x \in [0,2\pi]$, then the equation has a lot of other solutions as well, given by the trigonometric equations
$$\cos x = 0 \text{ and } \cos x = 1$$
for the first equation, the general solution is $x = n\pi + \frac{\pi}{2}$ and for the second equation, the general solution is $x = 2n\pi$. $\pi$ is not a solution as $\cos \pi = -1$. Infinite solutions exist because $\cos(x)$ is a periodic function. Have a look at the graph of $\cos x$ to clear things up:
If it's not mentioned that $x \in [0,2\pi]$, then the equation has a lot of other solutions as well, given by the trigonometric equations
$$\cos x = 0 \text{ and } \cos x = 1$$
for the first equation, the general solution is $x = n\pi + \frac{\pi}{2}$ and for the second equation, the general solution is $x = 2n\pi$. $\pi$ is not a solution as $\cos \pi = -1$. Infinite solutions exist because $\cos(x)$ is a periodic function. Have a look at the graph of $\cos x$ to clear things up: