Question

sin^2x+4cosx-3=0 AAAAAAAAAAAAAAAAA HELP

Answer 1

$sin^2x+4cosx-3=0\\\\1-cos^2x+4cosx-3=0\\\\cos^2x-4cosx+2=0\\\\1)\; \; (cosx)_1=\frac{4-\sqrt{16-8}}{2}=\frac{4-2\sqrt2}{2}=2-\sqrt2\; ,\\\\ x=\pm arccos(2-\sqrt2)+2\pi n,\; n\in Z\\\\2)\; \; (cosx)_2=2+\sqrt2\; \ \textgreater \ 1\; \; \to \; \; net\; reshenij$