Question

range of 1+3sin3x is ​

Answer 1

Answer:

ANSWER

y=(2−sin3x)1          --------- ( 1 )

Let f(x)=(2−sin3x)1

Domain  of f(x) is (−∞,∞)

From ( 1 ),

2−sin3x=y1

⇒ sin3x=2−y1

⇒  sin3x=y2y−1

⇒  3x=sin−1(y2y−1)         

⇒  x=31sin−1(y2y−1)          [ ∵  y=2−sin3x1∴y>0as−1≤sin3x≤1 ]

For x to be real

−1≤y2y−1≤1

⇒  −y≤2y−1≤y                         [ y>0 ]

⇒  2y−