Задания во вкладках...........

Решите задачу:

$cos \alpha =-\frac{\sqrt3}{2}\\\\sin \alpha =\pm \sqrt{1-cos^2 \alpha }=\pm \sqrt{1-\frac{3}{4}}=\pm \frac{1}{2}\\\\90^\circ \ \textless \ \alpha \ \textless \ 180^\circ \; \; \to \; \; sin \alpha =+\frac{1}{2}\\\\sin(30^\circ + \alpha )=sin30^\circ \cdot cos \alpha +cos30^\circ \cdot sin \alpha =\frac{1}{2}\cdot (-\frac{\sqrt3}{2})+\frac{\sqrt3}{2}\cdot \frac{1}{2}=0\\\\\\2)\; \; y=\sqrt{4-x}\\\\y^2=4-x\\\\x=4-y^2 \; \; \to \; \; y=4-x^2\; \; (obratnaya\; fynkciya)$