Найдите в градусах сумму корней уравнения :
sin x -2 cos^2 x= sin3x -1 на отрезке [-π/2;3π/2]
sin x -2 cos^2 x= sin3x -1 на отрезке [-π/2;3π/2]
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2sinxcos2x+cos2x=0
cos2x(2sinx+1)=0
cos2x=0⇒2x=π/2+πk⇒x=π/4+πk/2,k∈z
sinx=-1/2⇒x=-π/6+2πk U x=-5π/6+2πk,k∈z