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

Let $f(x) = \sqrt{3}\cos x + \sin x$.
Then $f(x) = 2 ({\sqrt{3} \over 2}\cos x + {1 \over 2} \sin x) = 2 (\sin { \pi \over 3} \cos x + \cos { \pi \over 3} \sin x) = 2 \sin (x+{ \pi \over 3})$, so to solve $f(x) = 1$, we need to find $x$ such that $\sin (x+{ \pi \over 3}) = {1 \over 2}$.
Since $\sin^{-1} (\{ {1 \over 2}\} ) = \{ { \pi \over 6}+2n\pi, { 5\pi \over 6}+2n\pi\}_n $, we see that the solutions are $\{ -{ \pi \over 6}+2n\pi, { \pi \over 2}+2n\pi\}_n $.
