Question

Тригонометрия 2sin^2x+√2sinx-10sinx-5√2=0

Answer 1

$2sin^2x+ (\sqrt{2} -10)sinx-5 \sqrt{2}=0;$
$sinx=t;t \in[-1;1];$
$2t^2+ (\sqrt{2} -10)t-5 \sqrt{2}=0;$
$D=(\sqrt{2} -10)^2+40 \sqrt{2}=2-20\sqrt{2}+100+40 \sqrt{2}=2+20\sqrt{2}+100=$
$=(\sqrt{2}10)^2;$
$t_1= \frac{-(\sqrt{2} -10)+(\sqrt{2}10)}{4}=5$ - посторонний корень
$t_2= \frac{-(\sqrt{2} -10)-(\sqrt{2}10)}{4}=- \frac{\sqrt{2}}{2};$
$sinx=- \frac{\sqrt{2}}{2};x=(-1)^narcsin(- \frac{\sqrt{2}}{2})+ \pi n;n \in Z;$
$x=(-1)^{n+1} \frac{ \pi }{4} + \pi n;n\in Z;$