Symmetry properties of $\sin$ and $\cos$. Why does $\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(-\frac{\pi}{2} - x\right)$? For a question such as: 

If $\sin(x) = 0.34$, find the value of $\cos\left(\frac{3\pi}{2} - x\right)$.

The solution says that:
\begin{align*}
\cos\left(\frac{3\pi}{2} - x\right) &= \cos\left(-\frac{\pi}{2} - x\right)\\
&= \cos\left(\frac{\pi}{2} + x\right)\\
&= -\sin(x)
\end{align*}
Why does $\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(-\frac{\pi}{2} - x\right)$?
Similarly, if $\cos(x) = 0.6$, find $\sin\left(\frac{3\pi}{2} + x\right)$, the solution says $\sin\left(\frac{3\pi}{2} + x\right)$ is equal to $\cos(\pi+x)$. How did they get $\cos(\pi+x)$?

Remember that cosine is $2\pi$ periodic this. This means that
$$\cos(x + 2n\pi) = \cos(x)$$
for any $x\in\mathbb{R}$ and for any integer $n\in\mathbb{Z}$. Basically, this says that if you can shift the graph of cosine left or right by $2\pi$ without changing it. Shifting it once gives you your desired equality since
$$\cos\left(-\frac{\pi}{2} -x\right)=\cos\left(2\pi - \frac{\pi}{2} - x\right)=\cos\left(\frac{3\pi}{2}-x\right)$$
Your second question is similar. Sine and cosine satisfy the shift equality
$$\sin\left(\frac{\pi}{2} + x\right)=\cos(x)$$
Graphically, this means that shifting the graph of sine by $\frac{\pi}{2}$ to the left gives the graph of cosine. Applying this property will give you your second equality.
