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.
 A: 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
$$
x=\pi-\sin^{-1}\!\left(\frac25\right)+\sin^{-1}\!\left(\frac45\right)
\implies\tan(x/2)=\frac{-3-\sqrt{21}}2
$$
A: Hint Rearranging gives (for $\frac{x}{2}$ in the image $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ of $\arctan$) that $$x = 2 \arctan t$$ (this is the celebrated if perhaps misnamed Weierstraß substitution, which has the convenient property of transforming rational functions in trigonometric functions of $x$ into rational functions in $t$, and which is particularly useful in computing antiderivatives of the former).
With this in hand, we can exploit the double-angle identity $\sin 2 u = 2 \sin u \cos u$ to write, e.g., $\sin x$ in terms of $t$:
$$\sin x = \sin 2 \arctan t = 2 \sin \arctan t \cos \arctan t .$$
Appealing to a reference triangle (appropriately labeling a right triangle with legs $t$ and $1$ and acute angle $x$, computing the length of the remaining side, and using the definition of the trigonometric functions) gives us (algebraic) simplifications:
$$\left\{
\begin{array}{rcl}
\sin \arctan t &=& \frac{t}{\sqrt{1 + t^2}} \\
\cos \arctan t &=& \frac{1}{\sqrt{1 + t^2}}\end{array}
\right.$$
Substituting gives
$$\sin x = 2 \left(\frac{t}{\sqrt{1 + t^2}}\right) \left(\frac{1}{\sqrt{1 + t^2}}\right) = \frac{2t}{1 + t^2}.$$
One can similarly derive a rational expression in $t$ for $\cos x$ and hence write the given equation in $x$ as a rational equation in $t$.
(NB as hinted above, this only detects the solutions with $\frac{x}{2}$ in an appropriate range. To find all of the solutions, one must use these in conjunction with the usual symmetries of the trigonometric functions themselves.)
A: I am posting only because I don't see the most obvious approach:
Since $t=\tan(x/2)=\frac{\sin(x/2)}{\cos(x/2)}$ and $\sin^2(x/2)+\cos^2(x/2)=1,$ solve for $\sin(x/2)=\frac{t}{\sqrt{1+t^2}}$ and $\cos(x/2)=\frac{1}{\sqrt{1+t^2}}$ (here I am assuming that $x/2\in [0,\pi/2)$-the other cases are similar).
Now you can use the formulas that give you the $\sin$ and $\cos$ of double angles, and you are done. 
A: Hint: $\sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}$,
$\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$
A: 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.
