Do enunciado, temos: $(\sqrt{3}\cos x )^2 = (1-\sin x)^2$ $\implies$ $3\cos^2 x = 1-2\sin x + \sin^2 x$ , temos: $4\sin^2 x - 2\sin x - 2 = 0$ $\implies$ $(\sin x - \frac{1}{4})^2 = (\frac{3}{4})^2$ , assim: $\sin x = \large{\frac{1 \pm 3}{4}}$ $=$ $\{-\frac{1}{2}, 1\}$ , portanto: $x = \{-\frac{\pi}{6} + 2k\pi \space \space \text{ou} \space \space x = \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z}\}$.