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(A) We know that $\operatorname{Sin}^{-1} x + \operatorname{Cos}^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.Given equation: $2 \operatorname{Cos}^{-1

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$0$

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(A) We know that $\operatorname{Sin}^{-1} x + \operatorname{Cos}^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.Given equation: $2 \operatorname{Cos}^{-1} x + \operatorname{Sin}^{-1} x = \frac{11 \pi}{6}$.We can rewrite this as: $\operatorname{Cos}^{-1} x + (\operatorname{Cos}^{-1} x + \operatorname{Sin}^{-1} x) = \frac{11 \pi}{6}$.Substituting the identity: $\operatorname{Cos}^{-1} x + \frac{\pi}{2} = \frac{11 \pi}{6}$.$\operatorname{Cos}^{-1} x = \frac{11 \pi}{6} - \frac{\pi}{2} = \frac{11 \pi - 3 \pi}{6} = \frac{8 \pi}{6} = \frac{4 \pi}{3}$.However,the range of $\operatorname{Cos}^{-1} x$ is $[0, \pi]$.Since $\frac{4 \pi}{3} > \pi$,there is no value of $x$ that satisfies this equation.Therefore,the number of solutions is $0$.

The number of solutions of the equation $2 \operatorname{Cos}^{-1} x + \operatorname{Sin}^{-1} x = \frac{11 \pi}{6}$ is

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