Question

sin2x=sin(3p/2+x)

Р.S. р-это пи

 

Answer 1

$<var>sin2x=sin(\frac{3\pi}{2}+x) \\ 2sinx*cosx=-cosx \\ 2sinx*cosx+cosx=0 \\ cosx(2sinx+1)=0 \\ 1) cosx=0 \ \ \ \ x=\frac{\pi}{2}+\pi k\\ 2)2sinx+1=0 \\ sinx=-\frac{1}2 \ \ \ \ x=(-1)^k*arcsin(-\frac1{2})+ \pi k \ \ \ \ x=(-1)^k*\frac{-\pi}{6}+ \pi k</var>$