Need help with $\arccos$ equation I have the equation
$$ \cos(2x + \frac{\pi}{9}) = 0.5$$
I know that in order to solve for $x\in \Bbb R$, I need to use
$$\arccos(0.5) = 2x + \frac{\pi}{9} $$
This yields
$$  2x + \frac{\pi}{9} =
\begin{cases}
\frac{\pi}{3} + 2k\pi,  & \text{Positive angle} \\
2 \pi - \frac{\pi}{3}+ 2k\pi, & \text{Negative angle}
\end{cases} $$
I would then subtract $\frac{\pi}{9}$ from both sides and get:
$$  2x  =
\begin{cases}
\frac{2\pi}{9} + 2k\pi,  & \text{Positive angle} \\
\frac{14\pi}{9}+ 2k\pi, & \text{Negative angle}
\end{cases} $$
However according to the handout the correct solution is:
$$  2x  =
\begin{cases}
\frac{4\pi}{9} + 2k\pi,  & \text{Positive angle} \\
\frac{16\pi}{9}+ 2k\pi, & \text{Negative angle}
\end{cases} $$
Can anyone help me?

Remember that $\;\cos x=\alpha\implies x=\pm\arccos x\;$. Besides this, we only need basic trigonometry to solve that equation:
$$\cos t=\frac12\iff t=\pm\frac\pi3+2k\pi\;,\;\;k\in\Bbb Z\implies$$
puting $\;t=2x+\frac\pi9\;$ we get
$$2x+\frac\pi9=\pm\frac\pi3+2k\pi\implies\begin{cases}2x=\cfrac{2\pi}9+2k\pi,&\text{(positive solution)}\\{}\\
2x=-\cfrac{4\pi}9+2k\pi=\cfrac{14\pi}9+2k\pi,&\text{(negative solution)}\end{cases}\;,\;k\in\Bbb Z$$
Thus you're right, the handout's solution is wrong.
