The number of solutions of the equation `"sin" x = [1+"sin" x] + [1 -"cos" x] "in" [0, 2pi]` is

Options

A. 0
B. 1
C. 2
D. 6

Answer

A

Explanation

We have,
`[1+"sin"x] = {{:(1", " 0 le x lt (pi)/(2)), (2", " x = (pi)/(2)), (1", " (pi)/(2) lt x le pi), (0", " pi lt x lt 2 pi), (1", " x = 2 pi):}`
`"and, "[1-"cos"x] = {{:(0", " 0 le x lt (pi)/(2)), (1", " (pi)/(2)le x lt pi), (2", " x=pi), (1", " pi lt x le (3pi)/(2)), (0", " (3pi)/(2) lt x le 2 pi):}`
`therefore [1+"sin"x] +[1-"cos"x] = {{:(1", " 0 le x lt (pi)/(2)), (3", "x = (pi)/(2)), (2", " (pi)/(2) lt x lt pi), (3", "x = pi), (1", " pi lt x le (3pi)/(2)),(0", " (3pi)/(2) lt x lt 2pi), (1", " x = 2pi):}`
Thus, the given equation reduces to
`"sin" x = 1, 0 le x lt (pi)/(2)`, which is not possible.
`"sin" x = 3, "at" x = (pi)/(2)`, which is absurd
`"sin"x = 2, (pi)/(2) lt x lt pi`, which is not possible
`"sin" x = 3"at" x = (pi)/(2)`, which is absurd
`"sin" x = 1, pi lt x le (3pi)/(2)`, which is impossible
`"sin" x = 0, (3pi)/(2) lt x lt 2 pi`, which is not possible
`"sin" x = 1, x = 2 pi`, which is absurd.
Hence, the given equation has no solution.