Question

Алгебра, тригонометрические уравнения. С чертежом окружности, если можно.
1) 2sin2x=tgx+ctgx
2) 3tg^2x-8cos^2x+1=0

Answer 1

$1)2Sin2x=tgx+Ctgx\\\\2Sin2x=\frac{Sinx}{Cosx}+\frac{Cosx}{Sinx}\\\\2Sin2x=\frac{Sin^{2}x+Cos^{2}x}{Sinx Cosx}\\\\2*2Sinx Cosx=\frac{1}{Sinx Cosx}\\\\4Sinx Cosx-\frac{1}{Sinx Cosx}=0\\\\Sinx Cosx = m\\\\4m-\frac{1}{m}=0 \ , \ m\neq0\\\\\frac{4m^{2}-1 }{m}=0\\\\4m^{2}-1=0\\\\m_{1} =\frac{1}{2}\\\\m_{2}=-\frac{1}{2}\\\\1)Sinx Cosx=\frac{1}{2}\\\\2Sinx Cosx=1\\\\Sin2x=1\\\\2x=\frac{\pi }{2}+2\pi n,n\in Z\\\\x=\frac{\pi }{4}+\pi n,n\in Z$

$2)Sinx Cosx=-\frac{1}{2}\\\\2Sinx Cosx=-1\\\\Sin2x=-1\\\\2x=-\frac{\pi }{2}+2\pi n,n\in Z\\\\x=-\frac{\pi }{4}+\pi n,n\in Z \\\\Otvet:\boxed{\pm\frac{\pi }{4}+\pi n,n\in Z}$

$2)3tg^{2}x-8Cos^{2}x+1=0\\\\(3tg^{2}x+1)-8Cos^{2}x=0\\\\\frac{3Sin^{2}x+Cos^{2}x }{Cos^{2}x } -8Cos^{2}x=0 \ , \ Cosx\neq 0\\\\3Sin^{2} x+Cos^{2}x-8Cos^{4}x=0\\\\3-3Cos^{2} x+Cos^{2}x-8Cos^{4}x=0\\\\8Cos^{4}x+2Cos^{2}x-3=0\\\\Cos^{2}x=m \ , \ m >0\\\\8m^{2}+2m-3=0\\\\D=2^{2}-4*8*(-3)=4+96=100=10^{2}\\\\m_{1} =\frac{-2+10}{16} =\frac{1}{2}\\\\m_{2}=\frac{-2-10}{16}=-\frac{3}{4} <0-neyd\\\\Cos^{2} x=\frac{1}{2}\\\\1)Cosx=\frac{\sqrt{2} }{2}$

$x=\pm\frac{\pi }{4}+2\pi n,n\in Z\\\\2)Cosx=-\frac{\sqrt{2} }{2}\\\\x=\pm\frac{3\pi }{4}+2\pi n,n\in Z$