5sin^2(x)-14sin(x)cos(x)-3cos^2(x)=2 Показать полное решение

$5sin ^{2} x-14sinxcosx-3cos ^{2} x=2$
$5sin ^{2} x-14sinxcosx-3cos ^{2} x-2(sin ^{2}x+cos ^{2}x) =0$
$5sin ^{2} x-14sinxcosx-3cos ^{2} x-2sin ^{2}x-2cos ^{2}x =0$
$3sin ^{2} x-14sinxcosx-5cos ^{2} x =0 |: cos ^{2} x,$ $(cos ^{2} x \neq 0)$

$3 \frac{sin ^{2}x }{cos ^{2}x } -14 \frac{sinx}{cosx} -5=0$

$3tg ^{2} x-14tgx-5=0$
$tgx=y$
$3y ^{2} -14y-5=0$
D = (-14)² - 4· 3·(-5) = 196 + 60 = 256

$\sqrt{D}=16$

$y_{1} = \frac{14+16}{6} = \frac{30}{6}=5$

$y_{2} = \frac{14-16}{6} = \frac{-2}{6} =- \frac{1}{3}$

$\left \{ {{tgx=5} \atop {tgx=- \frac{1}{3} }} \right.$

$x=arctg5+ \pi n,$ n∈Z
$x=-arctg \frac{1}{3} + \pi m,$ m∈Z