Inverse Trigonometric Functions Question 205
Question: The equation $ 2{{\cos }^{-1}}x+{{\sin }^{-1}}x=\frac{11\pi }{6} $ has
[AMU 1999]
Options:
A) No solution
B) Only one solution
C) Two solutions
D) Three solutions
Show Answer
Answer:
Correct Answer: A
Solution:
Given equation is $ 2{{\cos }^{-1}}x+{{\sin }^{-1}}x=\frac{11\pi }{6} $
Therefore $ {{\cos }^{-1}}x+({{\cos }^{-1}}x+{{\sin }^{-1}}x)=\frac{11\pi }{6} $
Therefore $ {{\cos }^{-1}}x+\frac{\pi }{2}=\frac{11\pi }{6} $
$ \Rightarrow {{\cos }^{-1}}x=4\pi /3 $ which is not possible as $ {{\cos }^{-1}}x\in [0,\pi ] $ .
BETA
