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.
$$\sqrt3\cos x=1-\sin x$$

Another approach :
\begin{align}
\sqrt3\cos x&=1-\sin x\\
\sqrt3\cos x+\sin x&=1\\
\frac{\sqrt3}{2}\cos x+\frac12\sin x&\stackrel{\color{red}{[1]}}=\frac12\\
\cos30^\circ\cos x+\sin30^\circ\sin x&\stackrel{\color{red}{[2]}}=\cos(60^\circ+360^\circ n)\\
\cos(x-30^\circ)&\stackrel{\color{red}{[3]}}=\cos(60^\circ+360^\circ n)\\
x-30^\circ&=60^\circ+360^\circ n\\
x&=90^\circ+360^\circ n\\
\end{align}

Notes :
$\color{red}{[1]}\;\;\;$Dividing both sides by $2$
$\color{red}{[2]}\;\;\;$Another solution is $\cos(360^\circ n-60^\circ)$
$\color{red}{[3]}\;\;\;\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y$
