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Решение задания смотри на фотографии
Разделим на√2:
1/√2 · sin3x - 1/√2 · cos3x = sinx
sin(3x - π/4) = sinx
sin(3x - π/4) - sinx = 0
2cos(2x - π/8) · sin(x - π/8) = 0
cos(2x - π/8) = 0 sin(x - π/8) = 0
2x - π/8 = π/2 + πn x - π/8 = πk
2x = 5π/8 + πn x = π/8 + πk
x = 5π/16 + πn/2
1)= -sin²a-tga-cosa*sina
2)= cosa*sina+ sin²a
3)= (-cosa*sina-sin²a)/(cos²a-sin²a)= (-sina)/cosa-sina