In triangle PQR,let angle P angle Q. If the | vedclass

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(C) Given the equation $4 \sin^3 x - 3 \sin x + a = 0$.Rearranging the terms,we get $a = 3 \sin x - 4 \sin^3 x$.Using the identity $\sin 3x = 3 \sin x

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$\frac{2\pi}{3}$

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(C) Given the equation $4 \sin^3 x - 3 \sin x + a = 0$.Rearranging the terms,we get $a = 3 \sin x - 4 \sin^3 x$.Using the identity $\sin 3x = 3 \sin x - 4 \sin^3 x$,the equation becomes $\sin 3x = a$.Since $\angle P$ and $\angle Q$ satisfy this equation,we have $\sin 3P = a$ and $\sin 3Q = a$.Thus,$\sin 3P = \sin 3Q$.Since $\angle P > \angle Q$,we consider the general solution $\sin 	heta = \sin \alpha \Rightarrow 	heta = n\pi + (-1)^n \alpha$.For $3P$ and $3Q$ to be angles in a triangle,$3P + 3Q = \pi$ (as $3P = \pi - 3Q$ is the only viable case for distinct angles $P$ and $Q$ in a triangle).Therefore,$3(P + Q) = \pi$,which implies $P + Q = \frac{\pi}{3}$.In $	riangle PQR$,we know $P + Q + R = \pi$.Substituting $P + Q = \frac{\pi}{3}$,we get $R = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

In $\triangle PQR$,let $\angle P > \angle Q$. If the radian measures of $\angle P$ and $\angle Q$ satisfy the equation $4 \sin^3 x - 3 \sin x + a = 0$ where $0 < a < 1$,then the radian measure of $\angle R$ is

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