Задание:
2-cosx=2sinx²x sin2x=cos3x
Решение:
2) sin (2x)=cos (3x) sin (2x)=cos (2x+x) sin (2x)=sos (2x)*cos (x) -sin (2x)*sin (x) sin (2x)=(1-2sin^2 (x)*cos (x) -2sin (x)*cos (x)*sin (x) sin (2x)=cos (x) -2sin^2 (x)*cos (x) -2sin^2 (x)*cos (x) sin (2x)=cos (x) -4sin^2 (x)*cos (x) 2sin (x)*cos (x)=cos (x) -4sin^2 (x)*cos (x) cos (x)*[2sin (x)+4sin^2 (x) -1]=01.cos (x)=0x=pi/2+pi*n 2. 4sin^2 (x)+2sin (x) -1=0 sin (x)=t 4t^2+2t-1=0 D=b^2-4ac=4+17=21 t1,2=(-b±sqrt (D) /2a t1=(-2+sqrt (21) /8 t2=(-2-sqrt (21) /8 sin (x)=(-2+sqrt (21) /8 x=(-1) ^n*arcsin (-2+sqrt (21) /8+pi*n sin (x)=(-2-sqrt (21) /8 x=(-1) ^n*arcsin (-2-sqrt (21) /8+pi*n
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