In a triangle ABC,angle A is greater than ang | vedclass

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(C) Given the equation $3\sin x - 4\sin^3 x = k$,we know that $\sin(3x) = k$.Since angles $A$ and $B$ satisfy this,we have $\sin(3A) = k$ and $\sin(3B

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

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(C) Given the equation $3\sin x - 4\sin^3 x = k$,we know that $\sin(3x) = k$.Since angles $A$ and $B$ satisfy this,we have $\sin(3A) = k$ and $\sin(3B) = k$.This implies $\sin(3A) - \sin(3B) = 0$,which gives $2\cos\left(\frac{3A+3B}{2}ight)\sin\left(\frac{3A-3B}{2}ight) = 0$.Since $A > B$,$\sin\left(\frac{3A-3B}{2}ight) 
eq 0$,so we must have $\cos\left(\frac{3A+3B}{2}ight) = 0$.Thus,$\frac{3(A+B)}{2} = \frac{\pi}{2} + n\pi$. For a triangle,$A+B If $\frac{3(A+B)}{2} = \frac{\pi}{2}$,then $A+B = \frac{\pi}{3}$,which gives $C = \pi - (A+B) = \frac{2\pi}{3}$.If $\frac{3(A+B)}{2} = \frac{3\pi}{2}$,then $A+B = \pi$,which is impossible for a triangle as $C$ would be $0$.

In a triangle $ABC$,angle $A$ is greater than angle $B$. If the measures of angles $A$ and $B$ satisfy the equation $3\sin x - 4\sin^3 x - k = 0$ for $0 < k < 1$,then the measure of angle $C$ is:

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