The equation 2cos^{-1}x + sin^{-1}x = frac{11 | vedclass

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(A) Given equation is $2\cos^{-1}x + \sin^{-1}x = \frac{11\pi}{6}$.We know that $\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.Rewr

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(A) Given equation is $2\cos^{-1}x + \sin^{-1}x = \frac{11\pi}{6}$.We know that $\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}$ for all $x \in [-1, 1]$.Rewriting the given equation: $\cos^{-1}x + (\cos^{-1}x + \sin^{-1}x) = \frac{11\pi}{6}$.Substituting the identity: $\cos^{-1}x + \frac{\pi}{2} = \frac{11\pi}{6}$.Solving for $\cos^{-1}x$: $\cos^{-1}x = \frac{11\pi}{6} - \frac{\pi}{2} = \frac{11\pi - 3\pi}{6} = \frac{8\pi}{6} = \frac{4\pi}{3}$.Since the range of $\cos^{-1}x$ is $[0, \pi]$,and $\frac{4\pi}{3} > \pi$,there is no value of $x$ that satisfies this equation.Therefore,the equation has no solution.

The equation $2\cos^{-1}x + \sin^{-1}x = \frac{11\pi}{6}$ has

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