The value of 'θ' satisfying sin⁡7θ=sin⁡4θ−sin⁡θ in 0<θ<π2 are

sin⁡7θ+sin⁡θ−sin⁡4θ=0⇒2sin⁡4θcos⁡3θ−sin⁡4θ=0⇒sin⁡4θ[2cos⁡3θ−1]=0sin4θ=0,cos3θ=124θ=π,3θ=π3⇒θ=π4,π9

The value of 'θ' satisfying sin⁡7θ=sin⁡4θ−sin⁡θ in 0<θ<π2 are

sin⁡7θ+sin⁡θ−sin⁡4θ=0⇒2sin⁡4θcos⁡3θ−sin⁡4θ=0⇒sin⁡4θ[2cos⁡3θ−1]=0sin4θ=0,cos3θ=124θ=π,3θ=π3⇒θ=π4,π9

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The value of ' $θ$ ' satisfying $\sin 7 θ = \sin 4 θ - \sin θ in 0 < θ < \frac{π}{2}$ are

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$\frac{π}{9}, \frac{π}{4}$

$\frac{π}{3}, \frac{π}{9}$

$\frac{π}{6}, \frac{π}{4}$

$\frac{π}{3}, \frac{π}{4}$

answer is A.

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Detailed Solution

$\begin{array}{l} \sin 7 θ + \sin θ - \sin 4 θ = 0 \\ \Rightarrow 2 \sin 4 θcos 3 θ - \sin 4 θ = 0 \\ \Rightarrow \sin 4 θ [2 \cos 3 θ - 1] = 0 \\ \sin 4 θ = 0, \cos 3 θ = \frac{1}{2} \\ 4 θ = π, 3 θ = \frac{π}{3} \Rightarrow θ = \frac{π}{4}, \frac{π}{9} \end{array}$