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dimka9828
@dimka9828
June 2022
1
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[tex](2sinx- \sqrt{3} )* \sqrt{3 x^{2} -7x+4} =0[/tex]
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m11m
Verified answer
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; +∞)
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Answers & Comments
Verified answer
1) 2sinx -√3 =02sinx = √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; +∞)