A) 1
B) 2
C) 0
D) \[\infty \]
Correct Answer: C
Solution :
\[{{\cos }^{-1}}x+{{\cos }^{-1}}(2x)=-\pi \] Þ \[{{\cos }^{-1}}2x=-\pi -{{\cos }^{-1}}x\] \[\Rightarrow 2x=\cos (\pi +{{\cos }^{-1}}x)\] Þ \[2x=\cos \pi (\cos {{\cos }^{-1}}x)-\sin \pi \sin ({{\cos }^{-1}}x)\] \[2x=-x\Rightarrow x=0\] But \[x=0\] does not satisfy the given equation. No solution will exist.
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