Question

B1 и С1 пожалуйста!! Срочно!!

Answer 1

$B1.\; \; \; cosx=\frac{2}{\sqrt{20}}\\\\1+tg^2x=\frac{1}{cos^2x}\; \; \Rightarrow \; \; tg^2x=\frac{1}{cos^2x}-1=(\frac{\sqrt{20}}{2})^2-1=\frac{20-4}{4}=\frac{16}{4}=4\\\\\frac{3\pi }{2}<x<2\pi \; \; \; \to \; \; \; tgx<0\\\\tgx=-\sqrt{4}=-2\\\\\\C1\; \; .\; \; \; 2sin^2x-\sqrt3\, sin2x=0\; \; \; \qquad (sin^2x=(sinx)^2\ne sinx^2=sin(x^2))\\\\2sin^2x-\sqrt3\cdot 2\, sinx\cdot cosx=0\\\\2\, sinx\cdot (sinx-\sqrt3\, cosx)=0\\\\a)\; \; sinx=0\; \; ,\; \; x=\pi n\; ,\; n\in Z\\\\b)\; \; sinx-\sqrt3\, cosx=0\; |:cosx\ne 0$

$tgx-\sqrt3=0\\\\tgx=\sqrt3\\\\x=arctg\sqrt3+\pi n=\frac{\pi}{3}+\pi n\; ,\; n\in Z\\\\Onvet:\; \; x=\pi n\; ;\; \; x=\frac{\pi}{3}+\pi n\; ,\; n\in Z\; .$