Question

2sinx cosx+4cos^2x=1

Answer 1

$2sinxcosx+4cos^2x=1 |:cos^2x neq 0 rightarrow x neq frac{ pi }{2} + pi k, k in Z\ 2tgx+4= frac{1}{cos^2x} \ 2tgx+4= 1+tg^2x\ tg^2x-2tgx-3=0\ tgx=t\ t^2-2t-3=0\ D = b^2-4ac=(-2)^2-4cdot1cdot(-3)=4+12=16\ t_1= frac{-b+ sqrt{D} }{2a}=frac{2+ sqrt{16}}{2}= frac{2+4}{2}=3\ t_2= frac{-b- sqrt{D} }{2a}=frac{2- sqrt{16}}{2}=frac{2-4}{2}=-1\ tgx=3\ x=arctg3+ pi n, nin Z\ tgx=-1\ x=arctg(-1)+ pi m, min Z\ x=-arctg(1)+ pi m, min Z\ x=- frac{ pi }{4} + pi m, min Z$