In order to make typing easier, I will let $t=x/2$. Replacing $\tan t$ by $\frac{\sin t}{\cos t}$, as you did, is a good general purpose strategy.
So the leftmost side is equal to
$$\frac{1-\frac{\sin t}{\cos t}}{1+\frac{\sin t}{\cos t}}.$$
The next step is semi-automatic. It seems sensible to bring top and bottom to the common denominator $\cos t$, and "cancel." We get
$$\frac{\cos t -\sin t}{\cos t+\sin t}.$$
The move after that is not obvious: Multiply top and bottom by $\cos t-\sin t$.
At the bottom we get $\cos^2 t-\sin^2 t$, which from a known double-angle formula we recognize as $\cos 2t$.
The top is $(\cos t-\sin t)^2$. Expand. We get $\cos^2 t-2\sin t\cos t+\sin^2 t$. The $\cos^2 t+\sin^2 t$ part is equal to $1$, and the $2\sin t\cos t$ part is equal to $\sin 2t$. So putting things together we find that the leftmost side is equal to
$$\frac{1-\sin 2t}{\cos 2t},$$
which is exactly what we want.
The second identity $\frac{1-\sin x}{\cos x}=\frac{\cos x}{1+\sin x}$ is much easier. In $\frac{1-\sin x}{\cos x}$, multiply top and bottom by $1+\sin x$. On top we now get $1-\sin^2 x$, which is $\cos^2 x$. At the bottom we get $\cos x(1+\sin x)$. Cancel a $\cos x$.
Remark: We can play a game of making the calculation more magic-seeming. The middle expression is equal to $\frac{1+\sin 2t}{\cos 2t}$. Replace the $1$ on top by $\cos^2 t+\sin^2 t$, and the $\sin t$ by $2\cos t\sin t$. Then on top we have $(\cos t+\sin t)^2$. Replace the $\cos 2t$ at the bottom by $\cos^2 t-\sin^2 t$, which is $(\cos t+\sin t)(\cos t -\sin t)$. Cancel the $\cos t+\sin t$, and we arrive at
$$\frac{\cos t+\sin t}{\cos t-\sin t}.$$
Now divide top and bottom by $\cos t$, and we get the desired $\frac{1+\tan t}{1-tan t}$.
Note that $$\cos x + \sin x = 0 \iff \cos x = -\sin x$$
Now, $\cos x$ cannot equal zero, since if it did, $\sin x = -1$ or $\sin x = 1$, in which case the given equation isn't satisfied.
So we can divide by $\cos x$ to get $$1 = \dfrac{-\sin x}{\cos x} = -\tan x \iff \tan x = -1$$
Solving for $x$ gives us the values $x = \dfrac {3\pi}4 + k\pi$, where $k$ is any integer.
Best Answer
Indicated Solution
We can derive the Weierstrass Substitution:
Using the tangent double angle formula: $$ \tan(x)=\frac{2t}{1-t^2}\tag{1} $$ Then writing $\sec^2(x)$ in terms of $\tan^2(x)$ $$ \begin{align} \sec^2(x) &=1+\tan^2(x)\\ &=1+\frac{4t^2}{1-2t^2+t^4}\\ &=\frac{1+2t^2+t^4}{1-2t^2+t^4}\\ &=\left(\frac{1+t^2}{1-t^2}\right)^2\tag{2} \end{align} $$ Therefore, checking sign of $\cos(x)$ vs $\tan(x/2)$: $$ \cos(x)=\frac{1-t^2}{1+t^2}\tag{3} $$ Multiplying $(1)$ and $(3)$ gives $$ \sin(x)=\frac{2t}{1+t^2}\tag{4} $$ Then, as mentioned in comments, we simply need to solve for $t=\tan(x/2)$: $$ 3\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}-4\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)}=2\tag{5} $$ which is simply a quadratic equation in $t$ giving $$ \tan(x/2)=t=\frac{-3\pm\sqrt{21}}2\tag{6} $$
Alternate Solution
Suppose $\theta$ is an angle so that $\sin(\theta)=\frac45$ and $\cos(\theta)=\frac35$; that is, $\theta=\sin^{-1}\!\left(\frac45\right)$.
Then, $$ \begin{align} \sin(x-\theta) &=\cos(\theta)\sin(x)-\sin(\theta)\cos(x)\\[3pt] &=\frac35\sin(x)-\frac45\cos(x)\\ &=\frac25 \end{align} $$ which gives the two solutions $$ x=\sin^{-1}\!\left(\frac25\right)+\sin^{-1}\!\left(\frac45\right) \implies\tan(x/2)=\frac{-3+\sqrt{21}}2 $$ and $$ x=\pi-\sin^{-1}\!\left(\frac25\right)+\sin^{-1}\!\left(\frac45\right) \implies\tan(x/2)=\frac{-3-\sqrt{21}}2 $$