Question

Just another cute problem:

Find the range of \(x\) for which \( \arccos \sqrt{x-1}- \arcsin \sqrt{2-x} =0 \) holds.

Answer 1

x => 1 And x <= 2
x \(\in\) [1,2]

Answer 2

Take cos of both sides:
\[\sqrt{x-1}=\cos({\arcsin{\sqrt{2-x}})}\]

Answer 3

@blockcolder: That's not how I did it.

@Ishaan94: How you did it?

Answer 4

cos(arcsin(x))=sqrt(1-x^2), so
\[\sqrt{x-1}=1-(2-x)=1+x\]
Then solve this for x.

Answer 5

\[\sqrt{x-1}=1-(2-x) \neq1+x\]

Answer 6

Oops. Typo. =))
\[\sqrt{x-1}=x-1\]
Only when x-1=0 or x-1=1 which means x=1 or 2.

Answer 7

Check again ^^

Answer 8

I got the domain for square-root part. And then checked if it agrees with the domain of inverse functions.

Answer 9

Interesting approach Ishaan. +1.