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The number of solutions of
$\sin θ + 2 \sin 2 θ + 3 \sin 3 θ + 4 \sin 4 θ = 10$ , $0 < θ < π$ is

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detailed solution

Correct option is A

Given equation is possible if sin⁡θ=sin⁡2θ=sin⁡3θ=sin⁡4θ=1for some θ,0<θ<π which is not possible .

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The number of solutions of sin⁡θ+2sin⁡2θ+3sin⁡3θ+4sin⁡4θ=10 , 0<θ<π is

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The number of solutions of
$\sin θ + 2 \sin 2 θ + 3 \sin 3 θ + 4 \sin 4 θ = 10$ , $0 < θ < π$ is