One of the solutions of the equation 8 sin^3 | vedclass

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(B) Given equation: $8 \sin^3 	heta - 7 \sin 	heta + \sqrt{3} \cos 	heta = 0$ Using the identity $4 \sin^3 	heta = 3 \sin 	heta - \sin 3 	heta$,

Vedclass · Accepted Answer

$\left(10^{\circ}, 20^{\circ}\right)$

Vedclass · Answer

(B) Given equation: $8 \sin^3 	heta - 7 \sin 	heta + \sqrt{3} \cos 	heta = 0$ Using the identity $4 \sin^3 	heta = 3 \sin 	heta - \sin 3 	heta$,we substitute: $2(3 \sin 	heta - \sin 3 	heta) - 7 \sin 	heta + \sqrt{3} \cos 	heta = 0$ $6 \sin 	heta - 2 \sin 3 	heta - 7 \sin 	heta + \sqrt{3} \cos 	heta = 0$ $\sqrt{3} \cos 	heta - \sin 	heta = 2 \sin 3 	heta$ Dividing by $2$: $\frac{\sqrt{3}}{2} \cos 	heta - \frac{1}{2} \sin 	heta = \sin 3 	heta$ $\sin(60^{\circ} - 	heta) = \sin 3 	heta$ $60^{\circ} - 	heta = 3 	heta \implies 4 	heta = 60^{\circ} \implies 	heta = 15^{\circ}$ Since $15^{\circ}$ lies in the interval $(10^{\circ}, 20^{\circ})$,the correct option is $B$.

One of the solutions of the equation $8 \sin^3 \theta - 7 \sin \theta + \sqrt{3} \cos \theta = 0$ lies in the interval

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