The possible values of θ belongs to 0,π such that sinθ+sin4θ+sin7θ=0 are
2π9,π4,4π9,π2,3π4,8π9
π4,5π12,π2,2π3,3π4,8π9
2π9,π4,π2,2π3,3π4,35π36
2π9,π4,π2,2π3,3π4,8π9
Explanation for the correct option:
Step 1: Solve sinθ+sin4θ+sin7θ=0
sinθ+sin4θ+sin7θ=0
sin4θ-3θ+sin4θ+sin4θ+3θ=0
[sin4θ-3θ+sin4θ+3θ]+sin4θ=0
2sin4θcos3θ+sin4θ=0 [∵sin(A-B)+sin(A+B)=2sinAcosB]
sin4θ2cos3θ+1=0
sin4θ=0,2cos3θ+1=0sin4θ=0,cos3θ=-12
Step 2: Solve for values of θ.
As we know that when sinθ=0⇒θ=nπ and when cosθ=cosα⇒θ=2nπ±α
sin4θ=0,cos3θ=cosπ34θ=nπ,3θ=(2n+1)π±π3θ=nπ4,θ=2n+13π±π9
Thus,0<nπ4<π;θ=2π9,4π9,8π9
Therefore,θ=2π9,π4,4π9,π2,3π4,8π9
Hence the correct option is option(A)
If fx=x-72x-27,x∈2,7 Then, the value of θ∈2,7 such that f'θ=0 is