<s>[INST] 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.
 [/INST] If you know the bisection formulas
$$
\left|\sin\frac{x}{2}\right|=\sqrt{\frac{1-\cos x}{2}}
\qquad
\left|\cos\frac{x}{2}\right|=\sqrt{\frac{1+\cos x}{2}}
$$
you can put them together finding
\begin{align}
\left|\tan\frac{x}{2}\right|=\sqrt{\frac{1-\cos x}{1+\cos x}}
&=\sqrt{\frac{1-\cos x}{1+\cos x}\frac{1-\cos x}{1-\cos x}}\\[6px]
&=\sqrt{\frac{(1-\cos x)^2}{\sin^2x}}\\[6px]
&=\biggl|\frac{1-\cos x}{\sin x}\biggr|\\[6px]
&=\frac{1-\cos x}{|\sin x|}
\end{align}
It's easy to check that $\tan(x/2)$ has the same sign as $\sin x$ for every $x$ where $\sin x\ne0$ (and so $\tan(x/2)$ is defined), which allows us to remove the absolute values:
$$
\tan\frac{x}{2}=\frac{1-\cos x}{\sin x}
$$
We can do similarly by using $1+\cos x$ in the second step instead of $1-\cos x$, which gives another neat formula:
$$
\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}
$$
Setting, for simplicity, $t=\tan(x/2)$, $X=\cos x$ and $Y=\sin x$, we can rewrite the equations as
$$
\begin{cases}
X+tY=1\\[6px]
tX-Y=-1
\end{cases}
$$
If we solve this for $X$ and $Y$, we arrive at
$$
X=\cos x=\frac{1-t^2}{1+t^2},\qquad
Y=\sin x=\frac{2t}{1+t^2}
$$

If you apply these to your equation, you get
$$
\frac{6t}{1+t^2}-\frac{4-4t^2}{1+t^2}=2
$$
that becomes a quadratic by removing the denominators:
$$
6t-4+4t^2=2+2t^2
$$
or, simplifying,
$$
t^2+3t-3=0
$$
Note there's a small issue: you have to verify that the values of $x$ for which $\tan(x/2)$ is not defined are or are not solutions of the equations. The check is easily done: $\tan(x/2)$ is not defined for $x=k\pi$, but
$$
3\sin(k\pi)-4\cos(k\pi)=\pm4\ne2
$$
If the equation is $3\sin x-4\cos x=4$, you get a family of solutions not covered by the $t$-substitution.
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