Помогите решить уравнения a)1+cos2x=sin2x; b)sin3xsinx+cos3xcosx=-1.
1) sin^2x+cos^2x+cos^2x-sin^2x = 2sinxcosx cos^2x = sinxcosx cosx(cosx-sinx) = 0 cosx = 0 => x = π/2 + πn cosx = sinx => 1 = tgx => x = π/4+πn 2)sin3xsinx+cos3xcosx=1 cos(3x-x) = 1 cos2x = 1 2x = 2πn x = πn
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спасибо большое