Question

Тригонометрические уровнения cosx=-√3/2 sin2x=-1 2ctg3x=-2 sinx< =1/2 cosx=> -√3/2

Answer 1

Cosx = -$\sqrt{3}$/2
x = pi - arccos $\sqrt{3}$/2  + 2pK
x = pi - pi/6 + 2pk
x1 = 5pi/6 + 2pk
x2= -pi + pi/6 + 2pk
x2 = -5pi/6 + 2pk

sin2x = - 1
2x = (-1)^karcsin(-1) + pK
2x= 3pi/2 + pk
x = 3pi/4 + pk/2

2ctg3x = -2
ctg3x = -2/2
ctg3x = -1
3x = arcctg(-1) + pk
3x = -arcctg1 + pk
3x = -pi/4 + pk
x = -pi/12 + pk/3

sinx <= 1/2 
x ∈ [ -pi - arcsin1/2 +2pk; arcsin1/2 + 2pk]
x ∈ [ -pi - pi/6 + 2pk; pi/6 + 2pk]
x ∈ [ -7pi/6 + 2pk; pi/6 + 2pk]

cosx =>-√3/2
x ∈ [arccos√3/2 + 2pk; -arccos √3/2 + 2pk]
x ∈ [ pi/6 + 2pk; -pi/6 + 2pk]