Consider
$$
\sin x -\sqrt 3 \cos x=1
$$
for $0\leq x \leq 2\pi$.
I solved it by converting $\sin x -\sqrt 3 \cos x$ to $2\sin(x-\pi/3)$ as follows
\begin{align*}
\sin x -\sqrt 3 \cos x &= r\sin(x-\theta)\\
&= r\sin x\cos\theta -r \cos x\sin\theta\\
\end{align*}
we have
\begin{align*}
r\cos \theta &= 1\\
r\sin \theta &=\sqrt 3
\end{align*}
where $r=2$ and $\theta=\frac{\pi}{3}$ are the solution.
So the equation becomes
$$
2\sin (x-\pi/3) =1
$$
and the solution are $x=\pi/2$ and $x=7\pi/6$.
Is there any other method to solve it?
Best Answer
Let me introduce the one and the only, the grandest substitution :
$t=\tan(\frac x 2)$
Now, $\sin(x)=\frac{2t}{1+t^2}$ and $\cos(x)=\frac{1-t^2}{1+t^2}$
Now put them back and solve the quadratic. Find value of $t$. Can you now find x?
Another method : Type the equation here