If 0 ≤ x < π 2 , then the number of values of x for which sin x - sin 2 x + sin 3 x = 0 , is:

If 0x<π2, then the number of values of x for which sinx-sin2x+sin3x=0, is:
  1. 4
  2. 3
  3. 2
  4. 1

Solution

sinx-sin2x+sin3x=0

sinx+sin3x-sin2x=0

2sin2 x .cosx-sin2x=0

sin2x=0 or cosx=12

2x=0 or x=π3

x=0 or  x=π3

 Number of solutions =2

Asked in: JEE Main 2019 (09 Jan Shift 2)

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