Question

Решите ! 2sin(x)+√2/2cos√2=0 2cosx-√3/2sinx+1=0

Answer 1

$2)\quad 2cosx-\frac{\sqrt3}{2}sinx+1=0\\\\2(cos^2\frac{x}{2}-sin^2\frac{x}{2})-\frac{\sqrt3}{2}\cdot 2sin\frac{x}{2}\cdot cos\frac{x}{2}+(sin^2\frac{x}{2}+cos^2\frac{x}{2})=0\\\\3cos^2\frac{x}{2}-sin^2\frac{x}{2}-\frac{\sqrt3}{2}sin\frac{x}{2}\cdot cos\frac{x}{2}=0\; |:cos^2\frac{x}{2}\ne 0\\\\3-tg^2\frac{x}{2}-\frac{\sqrt3}{2}tg\frac{x}{2}=0\; |\cdot (-2)\\\\2tg^2\frac{x}{2}+\sqrt3tg\frac{x}{2}-6=0\; \; ,\; \; t=tg\frac{x}{2}\\\\2t^2+\sqrt3t-6=0$

$D=3+48=51\; ,\; \; t_{1,2}=$$\frac{-\sqrt3\pm \sqrt{51}}{4}$  

$a)\; \; tgx= \frac{-\sqrt3+\sqrt{51}}{4} \; ,\; \; x=arctg \frac{\sqrt{51}-\sqrt3}{4}+\pi n\; ,\; n\in Z \\\\b)\; \; tgx= -\frac{\sqrt{51}+\sqrt3}{y} \; ,\; \; x=-arctg \frac{\sqrt{51}+\sqrt3}{4}+\pi k\; ,\; k\in Z$

$2)\quad 2sinx+\frac{\sqrt2}{2}cos\sqrt2=0\\\\sinx=-\frac{\sqrt2}{4}cos\sqrt2\approx -0,06\; ;\; \; \; \; \; -1\ \textless \ -0,06\ \textless \ 1\\\\x=(-1)^{n}arcsin(-\frac{\sqrt2}{4}cos\sqrt2)+\pi n=\\\\=(-1)^{n+1}arcsin(\frac{\sqrt2}{4}cos\sqrt2)+\pi n,\; n\in Z$