The possible values of theta in (0, pi) such | vedclass

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(D) Given equation: $\sin 	heta + \sin (4 	heta) + \sin (7 	heta) = 0$. Using the sum-to-product formula $\sin A + \sin B = 2 \sin \left( \frac{A+B

Vedclass · Accepted Answer

$\frac{2 \pi}{9}, \frac{\pi}{4}, \frac{4 \pi}{9}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{8 \pi}{9}$

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(D) Given equation: $\sin 	heta + \sin (4 	heta) + \sin (7 	heta) = 0$. Using the sum-to-product formula $\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} ight) \cos \left( \frac{A-B}{2} ight)$,we combine $\sin 	heta$ and $\sin (7 	heta)$: $2 \sin (4 	heta) \cos (3 	heta) + \sin (4 	heta) = 0$. Factor out $\sin (4 	heta)$: $\sin (4 	heta) (2 \cos (3 	heta) + 1) = 0$. This gives two cases: Case $1$: $\sin (4 	heta) = 0 \implies 4 	heta = n \pi \implies 	heta = \frac{n \pi}{4}$. For $	heta \in (0, \pi)$,the values are $\frac{\pi}{4}, \frac{2 \pi}{4} = \frac{\pi}{2}, \frac{3 \pi}{4}$. Case $2$: $2 \cos (3 	heta) + 1 = 0 \implies \cos (3 	heta) = -\frac{1}{2}$. $3 	heta = 2 n \pi \pm \frac{2 \pi}{3} \implies 	heta = \frac{2 n \pi}{3} \pm \frac{2 \pi}{9}$. For $	heta \in (0, \pi)$: If $n=0$,$	heta = \frac{2 \pi}{9}$. If $n=1$,$	heta = \frac{2 \pi}{3} - \frac{2 \pi}{9} = \frac{4 \pi}{9}$ and $	heta = \frac{2 \pi}{3} + \frac{2 \pi}{9} = \frac{8 \pi}{9}$. Combining all values,we get $	heta \in \{ \frac{2 \pi}{9}, \frac{\pi}{4}, \frac{4 \pi}{9}, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{8 \pi}{9} \}$.

The possible values of $\theta \in (0, \pi)$ such that $\sin \theta + \sin (4 \theta) + \sin (7 \theta) = 0$ are

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