1) Это однородное тригонометрическое уравнение второй степени .

Решим его делением обеих частей уравнения на Cos²x ,

(Cosx ≠ 0 ) .

[tex]\displaystyle\bf\\Sin^{2}x-3Cos^{2} x+2Sinx Cosx=0\\\\\\\frac{Sin^{2}x }{Cos^{2} x} +\frac{2Sinx Cosx}{Cos^{2}x } -\frac{3Cos^{2}x }{Cos^{2} x} =0\\\\\\tg^{2} x+2tgx-3=0\\\\tgx=m\\\\m^{2} +2m-3=0\\\\D=2^{2} -4\cdot(-3)=4+12=16=4^{2} \\\\\\m_{1} =\frac{-2-4}{2} =-3\\\\\\m_{2} =\frac{-2+4}{2} =1\\\\tgx=-3\\\\x=-arctg 3+\pi n,n\in Z\\\\tgx=1\\\\x=arctg 1+\pi n,n\in Z\\\\x=\frac{\pi }{4} +\pi n,n\in Z\\\\\\Otvet \ : \ -arctg3+\pi n \ \ ; \ \ \frac{\pi }{4} +\pi n,n\in Z[/tex]

[tex]\displaystyle\bf\\2)\\\\2Sin^{2} x-3Sinx-2=0\\\\Sinx=m \ , \ \ -1\leq m\leq 1\\\\2m^{2} -3m-2=0\\\\D=(-3)^{2} -4\cdot 2\cdot(-2)=9+16=25=5^{2} \\\\\\m_{1} =\frac{3-5}{4} =-\frac{1}{2} \\\\\\m_{2} =\frac{3+5}{4} =2 \ > 1 \ - \ ne \ podxodit\\\\\\Sinx=-\frac{1}{2} \\\\\\x=(-1)^{n} arc Sin\Big(-\frac{1}{2} \Big)+\pi n,n\in Z\\\\\\\boxed{x=(-1)^{n+1} \frac{\pi }{6} +\pi n,n\in Z}[/tex]