задания уровня С1 для 11 класса"А" (профильная подруппа)

Решить уравнения и найти корни, принадлежащие данному промежутку:
1) sin^2 x-4sinx+3=0 , [0; π]
2) 2cos^2 x+5cosx+3=0 , [-3π/2;-π/2]
3) sin^2 x+2sinx+1=0, [-π/2; π/2]
4) tg^2 x+2tgx-3=0, [ -2π;-π]


1) 2cos^2 x+3sinx=0, (-π/2;π/2)
2) sin^2 x+2cosx-2=0, (-π;0)
3) 2cos^2 x+2cosx+sin^2 x=0, (π/2;3π/2)


1) cos2x+3sinx=2, [-3π;π]
2) 3cos2x-5cosx=1, [0;2π]
3) cos2x+2=3cosx, [-5π/2;-π/2]
4) cos4x+cos2x=0, [-π;π/3]
5) cos4x-sin2x=0, [0;π]
6) cos6x-cos3x=0, [0;π]


1) sin^2 x+sinx=0, [π/2;3π/2]
2) cos^2 x-cosx=0, [-π/2;π]
3) sin2x∙cosx-2sinx=0, [-π;π]
4) 2sin2x+cosx+4sinx+1=0, [5π/2;7π/2]
5) 3sin2x-3cosx+2sinx-1=0, [-2π;-π]
6) 2sin2x-4cosx+3sinx-3=0, [π;5π/2]
7) sin2x=sin(π/2-x), [0;3π/2]


1) 2sin^2 x+sinx∙cosx-3cos^2 x=0, [π/2;3π/2]
2) 3sin^2 x+5sinx∙cosx+2cos^2 x=0, [-3π;-2π]
3) sin^2 x+3sinx∙cosx-4cos^2 x=0, [0;π]


1) 5sin^2 x-7sinx∙cosx+4cos^2 x=1, [π/2;5π/2]
2) 3cos^2 x-sin2x=0,5 , [-3π/2;0]
3) sin^2 x-sin2x-3cos^2 x=0, (-π/2;π/2)
4) 5sin^2 x-2sin2x+cos^2 x=0, (-π/2;π/2)