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Q.

If sin3α=4sinα sin(x+α) sin(x−α) then

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a

x=nπ±π3,n∈z

b

x=nπ±π6,n∈z

c

x=nπ±π2,n∈z

d

x=2nπ±π2,n∈z

answer is A.

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Detailed Solution

sin3α=4sinαsin(x+α)sin(x−α)⇒3sinα−4sin3α=4sinα(sin2x−sin2α)⇒3−4sin2α=4(sin2x−sin2α)⇒3=4sin2x⇒sin2x=34⇒sin2x=sin2π3 ⇒x=nπ±π3,n∈z

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