Since $f(x)=y$ and $g(x)=y$, you can set them equal to each other:

Insert $x$ into $f(x)=-x-1$ because it’s the simpler expression of the two. You could also insert $x$ into $g(x)$, but it would be more work.

The intersection between $f(x)$ and $g(x)$ is

Since $f(x)=y$ and $g(x)=y$, you can set them equal to each other:

Solving the quadratic equation with the quadratic formula:

so

Insert $x_1$ and $x_2$ into $g(x)=2x+3$ because it’s the simpler expression. You could also insert $x_1$ and $x_2$ into $f(x)$, but it would be more work.

Find the point of intersection between the graphs by setting them equal to each other:

You need to check that you didn’t miss any solutions since you divided by $\cos x$, which can be 0. You check this by looking at what happens if $\cos x=0$. Then $x=\frac{\pi }{2}$, which means $\sin x=1$, or $x=\frac{3\pi }{2}$, which means $\sin x=-1$. In both cases, $\sin x$ is different from $\cos x$, so they are not solutions.

There are infinitely many points of intersection, and the points have $x$-values equal to $x=\frac{\pi }{4}+n\pi$ for $n\in \mathbb{Z}$. You find the $y$-values by inserting the $x$-values into one of the functions, for example to $f(x)=\sin x$. Here, you have to be aware that even if $\tan x$ has a period of $\pi$, you have found two different angles on the unit circle, with $\pi$ radians between. This is why you get two different values when you put the result back into $f(x)=\sin x$:

These are two different $y$-values, $y=\frac{\sqrt{2}}{2}$ and $y=-\frac{\sqrt{2}}{2}$, each belonging to their respective angle on the unit circle. These angles repeat themselves with a period of $2\pi$. Together, you get two different points of intersection: