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Question

The possible value of θ(0,π) such that sin(θ)+sin(4θ)+sin(7θ)=0 are : ?

A
2π9,π4,4π9,π2,3π4,8π9
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B
π4,5π12,π2,2π3,3π4,8π9
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C
2π9,π4,π2,2π3,3π4,35π36
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D
2π9,π4,π2,2π3,3π4,8π9.
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Solution

The correct option is A 2π9,π4,4π9,π2,3π4,8π9
Here, 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(CD2)[sin(7θ)+sin(θ)]+sin(4θ)=02sin(7θ+θ2)cos(7θθ2)+sin(4θ)=02sin(4θ)cos(3θ)+sin(4θ)=0sin(4θ)[2cos(3θ)+1]=0sin(4θ)=0,2cos(3θ)+1=0
For,
sin(4θ)=04θ=sin1(0)4θ=π,2π,3π,4π,...θ=π4,π2,3π4,π,...
Here, θ(0,π)
And
2cos(3θ)+1=0cos(3θ)=123θ=cos1(12)3θ=ππ3,π+π3,3ππ3,3π+π3,...θ=2π9,4π9,8π9,...
Here, θ(0,π)
The possible values of $\theta =\dfrac { 2\pi }{ 9 } ,\dfrac { \pi }{ 4 } ,\dfrac { 4\pi }{ 9 } ,\dfrac { \pi }{ 2 } ,\dfrac { 3\pi }{ 4 } ,\dfrac { 8\pi }{ 9 } .$

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