I'm missing the right substitute $\sqrt3\cos x=1-\sin x$ Please show me how to solve the following equation for $x$.
I've tried multiple substitutes but can't seem to find the right one.
$$\sqrt3\cos x=1-\sin x$$

As our friends, has already showed everything,
$\sqrt3\cos x=1-\sin x$, now dividing by 2. We get,
$\frac{\sqrt3}{2}\cos x=\frac{1}{2}-\frac{1}{2}\sin x$
$\frac{\sqrt3}{2}\cos x+\frac{1}{2}\sin x=\frac{1}{2}$, or
$\cos(\pi/6)\cos x+\sin(\pi/6)\sin x=\frac{1}{2}$. So
$\cos(\pi/6-x) = \frac{1}{2}$. Or,
As $\cos(x)$ is an even function.
$(i)\pi/6-x = \pi/3+2\pi n$
$(ii)x-\pi/6 = \pi/3+2\pi n$
So,
$(i)x = \pi/6 -\pi/3 = -\pi/6+2\pi n$
$(ii)x = \pi/6 +\pi/3 = \pi/2+2\pi n$
