If $t = \tan (x/2)$, find expressions for $\sin x, \cos x$ in terms of $t$. Hence, solve the equation $3\sin x - 4\cos x = 2$. 
If $$t = \tan \frac{x}{2},$$ find expressions for $\sin x, \cos x$ in terms of $t$. Hence, solve the equation $$3\sin x - 4\cos x = 2.$$

Attempt:
I have been solving a lot of trig questions lately but this is different. I don't know how to approach this. I'm thinking of getting $\sin x$ and $\cos x$ from $\tan x$ and replacing in the equation but not sure how because of the $t$. Help please.

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
$$
x=\pi-\sin^{-1}\!\left(\frac25\right)+\sin^{-1}\!\left(\frac45\right)
\implies\tan(x/2)=\frac{-3-\sqrt{21}}2
$$
