What is the domain of #y=3/2 arcsin(2x+.5)#? Trigonometry Inverse Trigonometric Functions Graphing Inverse Trigonometric Functions 1 Answer turksvids Jan 17, 2018 #-3/4 <= x <= 1/4# Explanation: The domain of #sin^-1(u)# is #-1<=u<=1#, so for this problem we need to solve: #-1<=2x+1/2<=1# subtracting #1/2# from all parts: #-3/2 <= 2x <= 1/2# dividing through by 2: #-3/4 <= x <= 1/4# Answer link Related questions Question #b4e98 How do you graph inverse trigonometric functions? What is the domain and range of inverse trigonometric functions? How do you apply the domain, range, and quadrants to evaluate inverse trigonometric functions? Can the values of the special angles of the unit circle be applied to the inverse trigonometric... How do you graph #y = 2\sin^{-1}(2x)#? What is the domain and range for #y = xcos^-1[x]#? What is the domain and range for #y = 6sin^-1(4x)#? How do you find the domain and range for #y = 5arcsin(2cos(3x))#? How do you graph #y = Arctan(x/3) #? See all questions in Graphing Inverse Trigonometric Functions Impact of this question 2093 views around the world You can reuse this answer Creative Commons License