Question

2sin^2x+3 корень из 2 cos(3p/2+x)+2=0 [5p/2; 4p] решить

Answer 1

$2Sin^{2}x+3\sqrt{2}Cos(\frac{3\pi }{2}+x)+2=0\\\\2Sin^{2}x+3\sqrt{2}Sinx+2=0\\\\Sinx=m,-1\leq m\leq1\\\\2m^{2}+3\sqrt{2}m+2=0\\\\D=(3\sqrt{2})^{2}-4*2*2=18-16=2=(\sqrt{2})^{2}\\\\m_{1}=\frac{-3\sqrt{2}+\sqrt{2}}{4}= -\frac{\sqrt{2} }{2}\\\\m_{2}=\frac{-3\sqrt{2} -\sqrt{2}}{4}=-\sqrt{2}<-1$

$Sinx=-\frac{\sqrt{2}}{2}\\1)x=-\frac{\pi }{4}+2\pi n,n\in z \\\\n=2\\\\x=-\frac{\pi }{4}+\frac{16\pi }{4} =\frac{15\pi }{4}\\\\2)x=-\frac{3\pi }{4}+2\pi n,n\in z\\\\n=2\\\\x=-\frac{3\pi }{4}+\frac{16\pi }{4}=\frac{13\pi }{4}\\\\Otvet:\frac{13\pi }{4};\frac{15\pi }{4}$

a)

$-\frac{\pi }{4}+2\pi n,n\in z;-\frac{3\pi }{4}+2\pi n,n\in z\\\\b)\\\\\frac{13\pi }{4};\frac{15\pi }{4}$