If $\left(2+\sqrt{3}\right)\cos x=1-\sin x$ , then $x=$

${\left(2+\sqrt{3}\right)}^2{\cos }^{2}x={\left(1-\sin x\right)}^2$

$\left(7+4\sqrt{3}\right){\cos }^{2}x={\left(1-\sin x\right)}^2$

$\left(7+4\sqrt{3}\right)\left(1+\sin x\right)\left(1-\sin x\right)={\left(1-\sin x\right)}^2$

$\therefore \sin x=1$        (or)      $\frac{1+\sin x}{1-\sin x}=\frac{1}{7+4\sqrt{3}}$

$x=2n\pi +\frac{\pi }{2}$    (or)  $\therefore \sin x=-\frac{\sqrt{3}}{2}$

$x=2n\pi +\frac{\pi }{2}$  (or) $x=2n\pi -\frac{\pi }{3}$