Question
Download Solution PDFWhich of the following is the solution of the equation 4 sin x + 2√3 = 0 such that x ∈ [0, 2π]
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
If sin θ = sin α then θ = nπ + (- 1)n α, α ∈ [-π/2, π/2], n ∈ Z.
|
T-Ratios |
0° |
30° |
45° |
60° |
90° |
|
Sin |
0 |
1/2 |
1/√2 |
√3/2 |
1 |
|
Cos |
1 |
√3/2 |
1/√2 |
1/2 |
0 |
|
Tan |
0 |
1/√3 |
1 |
√3 |
Not defined |
Calculation:
Given: 4 sin x + 2√3 = 0 such that x ∈ [0, 2π]
⇒ 4 sinx = - 2√3
⇒ sin x = - √3/2
As we know that, sin x is negative in 3rd and 4th quadrant.
∵ x ∈ [0, 2π]
⇒ x = 4π/3, 5π/3
Hence, x = 4π/3, 5π/3 are the solution of the given equation.Last updated on Jun 18, 2025
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