Given that the inverse trigonometric function take principal value only. Then the number             of real value of x which satisfy  sin−13x5+sin−14x5=sin−1xis equal to:

Using sum of two inverse of sine functions and then remove  sine from both sides, we get  3x51-16x225+4x51-9x225=x ⇒3x25-16x2=25x-4x25-9x2⇒x=0 or ⇒325-9x2=25-425-9x2squaring on both sides⇒925-9x2=625-20025-9x2+1625-9x2Simplifying the above equation we get 20025-9x2=800⇒x2=1 ⇒x=±1 and x=0 ∴ Total number of solutions =3