Question

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

Answer 1

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


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