Difference between $\cos(x)$= positive and $\cos(x)$= negative I'm confusing myself here so need some clarification. If I was to work out the solutions to $\cos(x) = 1/\sqrt2$ , I know the solutions would be $\pi/4, 2\pi - \pi/4, 2\pi + \pi/4...$ etc.
What is the difference when working out the solution to $\cos(x) = -1/\sqrt2$. What values need to be added to what, and what are the solutions significance to the solutions when the value is positive (above) ?
 A: $\cos(x)=-\dfrac{1}{\sqrt{2}}$ implies that the solutions lie in the "negative" quadrants for cosine, i.e. the $2$nd and $3$rd quadrants.
So, for $x$, you would get:
$x=\big(\pi-\dfrac{\pi}{4}\big)$, $\big(\pi+\dfrac{\pi}{4}\big)$, ...
$\hspace{3 mm}=\dfrac{3\pi}{4}$, $\dfrac{5\pi}{4}$, ...
A: If $\cos x > 0$, then the terminal side of angle $x$ lies in the first quadrant, the fourth quadrant, or the positive $x$-axis.  If $\cos x < 0$, then the terminal side of angle $x$ lies in the second quadrant, the third quadrant, or the negative $x$-axis.
One solution of the equation $\cos x = -\dfrac{1}{\sqrt{2}}$ is 
$$x = \arccos\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4}$$

By symmetry, $\cos(-x) = \cos x$, so another solution is 
$$x = -\frac{3\pi}{4}$$
Any angle coterminal with one of those angles will have the same cosine.  Hence, the general solution is 
$$x = \pm \frac{3\pi}{4} + 2k\pi, k \in \mathbb{Z}$$
As the diagram indicates,
\begin{align*}
\cos(\pi + \theta) & = -\cos\theta\\
\cos(\pi - \theta) & = -\cos\theta
\end{align*}
which you can verify using the sum and difference of angle formulas for cosine.  You know that one solution of $\cos\theta = \dfrac{1}{\sqrt{2}}$ is
$$\theta = \arccos\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$$
so the solutions to the equation $\cos x = -\dfrac{1}{\sqrt{2}}$ include
\begin{align*}
x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\\
x = \pi + \frac{\pi}{4} = \frac{5\pi}{4}
\end{align*}
Observe that 
$$x = \frac{3\pi}{4} = \arccos\left(-\frac{1}{\sqrt{2}}\right)$$
and that 
$$x = \frac{5\pi}{4} = -\frac{3\pi}{4} + 2\pi$$
