Question

A)cos2x-5sinx+6=0 б)3cos2x=4-11cosx в)cos2x=6sinx-5=0 г)cos2x-5sinx+6=0

Answer 1

А)
$\displaystyle cos2x-5sinx+6=0\\\\(1-2sin^2x)-5sinx+6=0\\\\-2sin^2x-5sinx+7=0\\\\2sin^2x+5sinx-7=0\\\\D=25+56=81\\\\sinx= \frac{-5\pm 9}{4}\\\\ sinx=1; sin x \neq - \frac{7}{2}\\\\sinx=1\\\\x= \frac{ \pi }{2}+2 \pi n; n\in Z$

б)
$\displaystyle 3cos2x=4-11cosx\\\\3(2cos^2x-1)+11cosx-4=0\\\\6cos^2x+11cosx-7=0\\\\D=121+168=289=17^2\\\\cosx= \frac{-11\pm 17}{12}\\\\cosx= \frac{1}{2}; cosx \neq - \frac{7}{3}\\\\cosx= \frac{1}{2}\\\\x=\pm \frac{ \pi }{3}+2 \pi n; n\in Z$

в)
$\displaystyle cos2x+6sinx-5=0\\\\(1-2sin^2x)+6sinx-5=0\\\\-2sin^2x+6sinx-4=0\\\\2sin^2x-6sinx+4=0\\\\D=36-32=4\\\\sinx= \frac{6\pm 2}{4}\\\\sinx=1; sinx \neq 2\\\\sinx=1\\\\x= \frac{ \pi }{2}+2 \pi n; n\in Z$

г) смотри а)