Question

решите уравнение
а)2sinx-3cos^2x+2=0
б)5sin^2x-3sinxcosx-2cos^2x=0

Answer 1

$2sinx-3cos^2x+2=0\2sinx-3+3sin^2x+2=0\3sin^2x+2sinx-1=0\sinx=u\3u^2+2u-1=0\D:4+12=16\u=frac{-2pm4}{6}\\u_1=frac{1}{3}\sinx=frac{1}{3}\x=(-1)^narcsinfrac{1}{3}+pi n, ; nin Z;\\u_2=-1\sinx=-1\x=-frac{pi}{2}+2pi n, ; nin Z.$


$5sin^2x-3sinxcosx-2cos^2x=0|:cos^2x\5tg^2x-3tgx-2=0\tgx=u\5u^2-3u-2=0\D:9+40=49\u=frac{3pm7}{10}\\u_1=1\tgx=1\x=frac{pi}{4}+pi n, ; nin Z;\\u_2=-frac{2}{5}\tgx=-frac{2}{5}\x=-arctgfrac{2}{5}+pi n, ; nin Z;\\cosx neq 0\x neq frac{pi}{2}+pi n, ; nin Z.$