JEE Question

 if
  • a)
    θ = nπ + π/3, n ∈ I
  • b)
    θ = 2nπ ± π/3, n ∈ I
  • c)
    θ = 2nπ ± π/6, n ∈ I
  • d)
    θ = nπ +π/6, n ∈ I
Correct answer is option 'B'. Can you explain this answer?

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Answers

Krishna Nikkula
Nov 13, 2021
Cos3ø/2cos2ø-1=1/2
2cos3ø=2cos2ø-1
2cos2ø-2cos3ø=1
2(cos2ø-cos3ø)=1=> cos2ø-cos3ø=1/2
#cosC-cosD= 2sin(C+D/2) sin(D-C) /2
=>2sin(2ø+3(ø/2)) sin(3ø-2ø/2) =1/2
sin5(ø/2)sinø/2=1/2×1/2
=>sinø/2= 1/2
ø/2=nπ±π/6
ø=2nπ± π/3(0ption B)

Cos3ø/2cos2ø-1=1/22cos3ø=2cos2ø-12cos2ø-2cos3ø=12(cos2ø-cos3ø)=1=> cos2ø-cos3ø=1/2#cosC-cosD= 2sin(C+D/2) sin(D-C) /2=>2sin(2ø+3(ø/2)) sin(3ø-2ø/2) =1/2sin5(ø/2)sinø/2=1/2×1/2=>sinø/2= 1/2ø/2=nπ±π/6ø=2nπ± π/3(0ption B)

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Cos3ø/2cos2ø-1=1/22cos3ø=2cos2ø-12cos2ø-2cos3ø=12(cos2ø-cos3ø)=1=> cos2ø-cos3ø=1/2#cosC-cosD= 2sin(C+D/2) sin(D-C) /2=>2sin(2ø+3(ø/2)) sin(3ø-2ø/2) =1/2sin5(ø/2)sinø/2=1/2×1/2=>sinø/2= 1/2ø/2=nπ±π/6ø=2nπ± π/3(0ption B)