Q48 / 90 · JEE Main 2021 (16 Mar Shift 1) · Mathematics

The number of roots of the equation, 81sin2x+81cos2x=30 in the interval 0,π is equal to :

  • A
    3
  • Correct answer
    B
    4
  • C
    8
  • D
    2

Solution

81sin2x+81cos2x=30

81sin2x+81181sin2x=30

81sin2x=t

t+81t=30

t2-30t+81=0

t-3t-27=0

81sin2x=31 or  81sin2x=33

34sin2x=31  or   34sin2x=33

sin2x=14  or  sin2x=34

Figure  

Hence, total 4 solution.

Hence, the correct answer is option: 'B'.

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