Задание:
cos2x+cosx+1/2sinx+V3=0
Решение:
) 1/2cos2x+V3/2sin2x=cos^2x+sin^2x 1/2cos^2x-1/2sin^2x+V3sinxcosx=cos^2x+sin^2x 1,5sin^2x-V3sinxcosx+0,5cos^2x=0, делим на cos^2x 1,5tg^2x-V3tgx+0,5=0 3tg^2x-2V3tgx+1=0 y=tgx 3y^2-2V3y+1=0 (V3y-1) ^2=0 V3y=1 y=1/V3=V3/3 tgx=V3/3-> x=pi/6+pi*n 2) cos^2x+sin2x-3sin^2x=0 3sin^2x-2sinxcosx-cos^2x=0, делим на cos^2x: 3tg^2x-2tgx-1=0 y=tgx 3y^2-2y-1=0 y1=-1/3 y2=1 x1=arctg (-1/3)+pi*n x2=pi/4+pi*n 3) 1+cos4x=cos2x (cos2x) ^2+(sin2x) ^2+(cos2x) ^2- (sin2x) ^2)=cos2x 2 (cos2x) ^2-cos2x=0 cos2x (2cos2x-1)=0 cos2x=0->2x=2pi*n->x1=pi*n 2cos2x=1->cos2x=1/2->2x=+-pi/6+2pi*n->x2=+-pi/12+pi*n 4) sin2x/ (1-cosx)=2sinx 2sinxcosx=2sinx (1-cosx) sinxcosx=sinx-sinxcosx 2sinxcosx-sinx=0 sinx (2cosx-1)=0 sinx=0->x1=pi*n 2cosx=1->cosx=1/2->x2=+-pi/6+2pi*n
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