find the domain of

sin inverse [sec x]   where [.] represents the greatest integer functions..

yup.........it is a null set i.e da function does nt exists as da value of  [sec x ] never lies b/w [-1 , 1]

& sin inverse x is defined only fr x = [-1 ,1]

EDIT-lozzzz ......made a mistake secx =-1 so dat  [sec x ] =[-1,1]

so domain aint null

the domain of sin ^-1 x is [-1,1]

sec x lies between [-infinity,-1] union [1,infinity]

sec x is equal to -1 at (2n+1)  ,so [sec x]=-1 at this value

[sec x] =1 means sec x lies between [1,2)

that happens  when x lies between [0, /3) ,general eq. is [2n ,2n +-  /3)

so the final answer is (2n+1)  union [2n ,2n +-  /3)

the inverse exists only if

[secx] = -1 & +1

so

secx belongs to [-1,0) U [1,2)

in [-1,0) only {-1} is possibl--------so x= (2n+1)pi

in [1,2)-----x takes values from [2n pi , 2n pi +- sec inverse2)

            --------------[2n pi , 2n pi +- pi/3)

finally take the union....

hey y r u guyz missing the +- .........sec follows cos .......nd so wen u take gen solution u'll have( 2npi +-  alpha)

Hope that Helps!

ya, sorry abt that .i forgot that in cos you have to take the +- part in the general eqn.

hi @akshaykhare

i have also got the same answer..........i used +- jus to make the sign omitters kno their mistakes.......

i stated [2n pi , 2npi + - pi/3)

unstead of my +- u've used the conventional method-------( 2n pi - pi/3, 2npi + pi/3)--------

we've got the same answer!