Can someone please check these trigonometric problems about the function $g(x) = \cos x$? $g (x) = \cos x$
A) What is the $y$ intercept? 
I got $(0,1)$.
B) For what numbers $x$ is $-\pi \leq x \leq \pi$ find where the graph is decreasing?
I got $0<x<\pi$.
C) What is the absolute minimum?
I got $-1$.
D) For what numbers $x$, $0 \leq x \leq 2\pi$, does the function $=0$?
I got $x= \pi/2$, $x=3\pi/2$, and $x=5\pi/2$.
E) For what numbers $x$, $-2\pi \leq x \leq 2\pi$ does $g(x)=1$? Where does it equal $-1$?
I got it equals $1$ when $x=-2\pi$, $0$, and $2\pi$.
It equals $-1$ when $x= \pi, -\pi$.
F) For what numbers $x$, $-2\pi \leq x \leq 2\pi$ does $g(x)= \sqrt{3}/2$?
I got $k\pi/2$ and $k$ is an integer, and that's all.
G) What are the $x$ intercepts of $g$?
I got $(-\pi/2,0)$, $(\pi/2,0)$, $(3\pi/2, 0)$, $(-3\pi/2, 0)$, $(5\pi/2,0)$ $(-5\pi/2,0)$. 
 A: Your answers to A, C, and E are correct.
$g(x) = \cos x$

(b) In the interval $-\pi \leq x \leq \pi$, where is the graph decreasing?

A function $f$ is (strictly) decreasing on an interval $I$ if for each $x_1, x_2 \in I$, with $x_1 < x_2$, $f(x_1) > f(x_2)$.  
For the function $g(x) = \cos x$ restricted to the domain $[-\pi, \pi]$, the function is decreasing in the interval $[0, \pi] = \{x \in \mathbb{R} \mid 0 \leq x \leq \pi\}$.

(d) For which numbers $x$, $0 \leq x \leq 2\pi$, does $g(x) = 0$?

You answers $\pi/2$ and $3\pi/2$ are both correct.  While $\cos\left(\frac{5\pi}{2}\right) = 0$, $\frac{5\pi}{2} > \frac{4\pi}{2} =  2\pi$, so $5\pi/2$ is not a valid solution in this interval.

(f)  For which numbers $x$, $-2\pi \leq x \leq 2\pi$, does $g(x) = \dfrac{\sqrt{3}}{2}$?

A particular solution of the equation 
$$\cos x = \frac{\sqrt{3}}{2}$$
is 
$$x = \arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$$
For the others, we can use symmetry. 

Two angles have the same cosine if the $x$-coordinates of the points where there terminal sides intersect the unit circle are equal.  Hence, $\cos\theta = \cos\varphi$ if $\varphi = \theta$ or $\varphi = -\theta$.  
Thus, 
$$x = -\frac{\pi}{6}$$ 
is a solution.  
Coterminal angles have the same cosine.  Hence, $\cos\theta = \cos\varphi$ if 
$$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$
or 
$$\varphi = -\theta + 2m\pi, m \in \mathbb{Z}$$
We can express the two equations above in the form 
$$\varphi = \pm\theta + 2n\pi, n \in \mathbb{Z}$$
In this problem, we know that $x = \frac{\pi}{6}$ is a solution.  Hence, any solution to the equation $\cos x = \frac{\sqrt{3}}{2}$ has the form 
$$x = \pm \frac{\pi}{6} + 2n\pi, n \in \mathbb{Z}$$
In the interval $[-2\pi, 2\pi]$, the only solutions are 
\begin{align*}
x & = \frac{\pi}{6}\\
x & = -\frac{\pi}{6}\\
x & = \frac{\pi}{6} - 2\pi = -\frac{11\pi}{6}\\
x & = -\frac{\pi}{6} + 2\pi = \frac{11\pi}{6}
\end{align*}

(g) What are the $x$-intercepts of $g(x) = \cos x$?

All of the numbers you listed are valid $x$-intercepts.  However, no interval was specified.  Therefore, the solution set consists of all angles whose terminal side lies on the positive or negative $y$-axis.  The angles whose terminal sides lie on the positive $y$-axis are 
$$x = \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z}$$
The angles whose terminal sides lie on the negative $y$-axis are 
$$x = -\frac{\pi}{2} + 2m\pi, m \in \mathbb{Z}$$
Notice that we could also use symmetry.  A particular solution of the equation $$\cos x = 0$$
is 
$$x = \arccos(0) = \frac{\pi}{2}$$
By symmetry,
$$x = -\frac{\pi}{2}$$ 
is also a solution.  Since any angle coterminal with these angles is a solution, the general solution is
$$x = \pm \frac{\pi}{2} + 2n\pi, n \in \mathbb{Z}$$
