Question

Решить уравнения 1)4sin^2x+11sinx+7=0 2)8sin^2-14cosx+1=0 3)2sin^2x+9sinxcosx+9cos^2x=0 вообще не разбираюсь в этом.

Answer 1

$4sin^2x+11sinx+7=0\\sinx=u\\4u^2+11u+7=0\\D:121-112=9\\u=\frac{-11\pm3}{8}\\\\u_1=-1\\sinx=-1\\x=-\frac{\pi}{2}+2\pi n, \; n\in Z;\\\\u_2=-\frac{7}{4}\\sinx \neq -\frac{7}{4}, \; sinx\in [-1;1].$

$8sin^2x-14cosx+1=0\\8-8cos^2x-14cosx+1=0|*(-1)\\8cos^2x+14cosx-9=0\\cosx=u\\8u^2+14u-9=0\\D:196+288=484\\u=\frac{-14\pm22}{16}=\frac{-7\pm 11}{8}\\\\u_1=\frac{1}{2}\\cosx=\frac{1}{2}\\x=\pm\frac{\pi}{3}+2\pi n, \; n\in Z;\\\\u_2=-\frac{9}{4}\\cosx \neq -\frac{9}{4}, \; cosx\in [-1;1].$

$2sin^2x+9sinxcosx+9cos^2x=0|cos^2x\\2tg^2x+9tgx+9=0\\tgx=u\\2u^2+9u+9=0\\D:81-72=9\\u=\frac{-9\pm 3}{4}\\\\u_1=-\frac{3}{2}\\tgx=-\frac{3}{2}\\x=-arctga\frac{3}{2}+\pi n, \; n\in Z;\\\\u_2=-3\\tgx=-3\\x=-arctg3+\pi n, \; n\in Z.$