Question

Решить уравнение

$4+2cos^2x=7sin2x$

Answer 1

$4+2cos^2x=7sin2x\\4(underbrace {sin^2x+cos^2x}_{1})+2cos^2x=7cdot underbrace {2sinxcdot cosx}_{sin2x}\\4sin^2x-14cdot sinxcdot cosx+6cos^2x=0; |:cos^2xne 0\\4tg^2x-14tgx+6=0; |:2\\2tg^2x-7tgx+3=0\\t=tgx; ,; ; ; 2t^2-7t+3=0; ,; ; D=49-24=25; ,\\t_1= frac{7-5}{4} =frac{1}{2} ; ,; ; t_2= frac{7+5}{4}=3\\a); ; tgx=frac{1}{2} \\underline {x=arctg frac{1}{2}+pi n,; nin Z} \\b); ; tgx=3\\underline {x=arctg3+pi k,; kin Z}$