If 3 sin 2 theta = 2 sin 3 theta and 0

Question

(D) Given the equation: $3 \sin 2 	heta = 2 \sin 3 	heta$. Using the identities $\sin 2 	heta = 2 \sin 	heta \cos 	heta$ and $\sin 3 	heta = 3 \

Vedclass · Accepted Answer

$\frac{\sqrt{15}}{4}$

Vedclass · Answer

(D) Given the equation: $3 \sin 2 	heta = 2 \sin 3 	heta$. Using the identities $\sin 2 	heta = 2 \sin 	heta \cos 	heta$ and $\sin 3 	heta = 3 \sin 	heta - 4 \sin^3 	heta$: $3(2 \sin 	heta \cos 	heta) = 2(3 \sin 	heta - 4 \sin^3 	heta)$. $6 \sin 	heta \cos 	heta = 6 \sin 	heta - 8 \sin^3 	heta$. Since $0 $3 \cos 	heta = 3 - 4 \sin^2 	heta$. Substitute $\sin^2 	heta = 1 - \cos^2 	heta$: $3 \cos 	heta = 3 - 4(1 - \cos^2 	heta) = 3 - 4 + 4 \cos^2 	heta$. $4 \cos^2 	heta - 3 \cos 	heta - 1 = 0$. Factoring the quadratic: $(4 \cos 	heta + 1)(\cos 	heta - 1) = 0$. This gives $\cos 	heta = 1$ (which implies $	heta = 0$,not in range) or $\cos 	heta = -\frac{1}{4}$. Using $\sin^2 	heta = 1 - \cos^2 	heta$: $\sin^2 	heta = 1 - (-\frac{1}{4})^2 = 1 - \frac{1}{16} = \frac{15}{16}$. Therefore,$\sin 	heta = \frac{\sqrt{15}}{4}$.

If $3 \sin 2 \theta = 2 \sin 3 \theta$ and $0 < \theta < \pi$,then the value of $\sin \theta$ is equal to

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