Ответы на вопрос:
1+ cos4x = cos2x1 + 2cos²2x -1 = cos2x 2cos²2x-cos2x=0 cos2x(2cos2x-1)=0 [cos2x=0 [cos2x=1/2 \\в объединение [2x=π/2 + πn n∈z [2x=±π/3 + πk k∈z [x=π/4 + πn/2 n∈z [x=±π/6 + πk/2 k∈z 4sin^2 x - 4sin x + 1 = 0пусть sin x=t, |t|≤1 4t²-4t+1=0 (2t-1)²=0 t=1/2 sinx=1/2 [x=π/6 + 2πn n∈z [x=5π/6 + 2πk k∈z
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