The correct option is
A 2π9,π4,4π9,π2,3π4,8π9Here, given
θ∈(0,π)and sin(θ)+sin(4θ)+sin(7θ)=0⇒[sin(7θ)+sin(θ)]+sin(4θ)=0
We know that,
sinC+sinD=2sin(C+D2)cos(C−D2)∴[sin(7θ)+sin(θ)]+sin(4θ)=0⇒2sin(7θ+θ2)cos(7θ−θ2)+sin(4θ)=0⇒2sin(4θ)cos(3θ)+sin(4θ)=0⇒sin(4θ)[2cos(3θ)+1]=0⇒sin(4θ)=0,2cos(3θ)+1=0
For,
sin(4θ)=0⇒4θ=sin−1(0)⇒4θ=π,2π,3π,4π,...⇒θ=π4,π2,3π4,π,...
Here, θ∈(0,π)
And
2cos(3θ)+1=0⇒cos(3θ)=−12⇒3θ=cos−1(−12)⇒3θ=π−π3,π+π3,3π−π3,3π+π3,...⇒θ=2π9,4π9,8π9,...
Here, θ∈(0,π)
∴ The possible values of $\theta =\dfrac { 2\pi }{ 9 } ,\dfrac { \pi }{ 4 } ,\dfrac { 4\pi }{ 9 } ,\dfrac { \pi }{ 2 } ,\dfrac { 3\pi }{ 4 } ,\dfrac { 8\pi }{ 9 } .$