Ответ(ы) на вопрос:
Sin^2x + cos^2 2x + sin^2 3x = 3/2
1 - cos2x + 1 + cos4x + 1 - cos6x = 3
cos4x - cos6x - cos2x = 0
cos 4x - (cos6x + cos 2x) = 0
cos4x - 2cos4xsin2x = 0;
cos4x(1 - 2sin2x) = 0;
cos4x = 0; 4x = π/2 + πn,n ∈ Z; x = π/8 + πn/4, n ∈ Z;
sin2x = 1/2 ; 2x = π/6 + πn, n ∈ Z; x = π/12 + πn/2, n ∈ Z;
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