1) 2sinx -√3 =0
2sinx = √3
sinx = √3
2
x= (-1)^k * (π/3) +πk, k∈Z
2) √(3x² -7x+4) =0
ОДЗ: 3x²-7x+4 ≥0
3x²-7x+4 =0
D= 49-48=1
x₁ = 7-1 = 1
6
x₂ = 7+1 =8/6 = 4/3 = 1 ¹/₃
6
+ - +
-------- 1 ----------- 1 ¹/₃ --------------
\\\\\\\\\\\ \\\\\\\\\\\\\\\\\
x∈(-∞; 1]U[1 ¹/₃; +∞)
При к=0
х=(-1)⁰ * (π/3) = π/3 = 3,14/3 ≈ 1,047 - не подходит по ОДЗ.
Ответ: 1; 1 ¹/₃ ; (-1)^k * (π/3) + πk, k∈(-∞; 0)U(0; +∞)