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