Is there any other method to solve $\sin x -\sqrt 3 \cos x=1$?

Question

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?

Answer 1

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

Answer 2

Squaring you have:

$$\sin^2 x + 3\cos^2x - 2\sqrt{3}\sin x\cos x = 1$$

So, $\cos^2 x = \sqrt{3}\sin x\cos x$

i.e., either $\cos x = 0$, in which case $x = \pi/2$ (since $3\pi/2$ is not a solution) or $\tan x= 1/\sqrt{3}$, in which case $x = 7\pi/6$

Answer 3

You can convert it directly:

$$\sin x -\sqrt 3 \cos x=\sin x -\tan(\frac{\pi}{3}) \cos x=\sin x -\frac{\sin(\frac{\pi}{3})}{\cos(\frac{\pi}{3})} \cos x\\ =\frac{\sin x\cos(\frac{\pi}{3}) -\sin(\frac{\pi}{3})\cos x}{\cos(\frac{\pi}{3})} =\frac{\sin (x-\frac{\pi}{3})}{\cos(\frac{\pi}{3})} $$

In general, when dealing with $\sin(x)\pm A \cos(x)$, to convert it just write $A=\tan(\theta)$.

Answer 4

We know that $\cos{\pi\over3}={1\over2}$ and $\sin{\pi\over3}={\sqrt3\over2}$, so

$$\begin{align} \sin x-\sqrt3\cos x=1&\implies{1\over2}\sin x-{\sqrt3\over2}\cos x={1\over2}\\ &\implies\sin x\cos{\pi\over3}-\cos x\sin{\pi\over3}={1\over2}\\ &\implies\sin(x-{\pi\over3})={1\over2}\\ &\implies x-{\pi\over3}={\pi\over6}\quad\text{or}\quad x-{\pi\over3}=\pi-{\pi\over6}\\ &\implies x={\pi\over2}\quad\text{or}\quad x={7\pi\over6} \end{align}$$