1) а)arcCos(-√2/2)+arcsin(-1); б)arcCos 1/2 - arctg 1/√3 2) a)sin3x=√2/2

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вопрос

a)
1) а) arccos(-√2/2) = -π/4
arcsin(-1) = -π/2
-π/4 + (-π/2) = -3π/4

б) arccos(1/2) = π/3
arctg(1/√3) = π/6
π/3 - π/6 = π/6

2) a) sin3x = √2/2
Since sin(π/4) = √2/2, this means that 3x = π/4 + 2πk or 3x = 3π/4 + 2πk, where k is an integer.
Therefore, x = π/12 + (2/3)πk or x = π/4 + (2/3)πk, where k is an integer.