Question

решите уравнение: sin2x + 2cos^2x=0

Answer 1

$sin2x+2cos^2x=0\\ 2sinx*cosx+2cos^2x=0\\ 2sinx*cosx=-2cos^2x\\ 2cosx(cosx+sinx)=0\\ cosx=0\\ x=\frac{\pi}{2}+\pi\*k\\ cosx=-sinx\\ tgx=-1\\ x=-\frac{\pi}{4}+\pi\*k$

Answer 2

$<var>sin2x + 2cos^2x=0 \\ \\ 2sinxcosx+2cos^2x=0 \\ \\ 2cosx(sinx+cosx)=0 \\ \\ cosx=0 \\ \\ x_1=\frac{\pi}{2}+\pi n,\ \ \ \ \ n \in Z \\ \\ sinx+cosx=0 \\ \\ tgx+1=0 \\ \\ tgx=-1 \\ \\ x_2=-\frac{\pi}{4}+\pi n,\ \ \ \ \ n \in Z</var>$