Q42 / 71 · NTA Abhyas (JEE Main) 2020 · Mathematics · Inverse Trigonometric Functions

For x∈0,π2, if cos-1721+cos⁡2x+sin2 x-48cos2 xsin⁡x=x-cos-1kcos⁡x, 
then the value of k is equal to

  • A
    1
  • B
    5
  • Correct answer
    C
    7
  • D
    14

Solution

y=cos-1721+cos⁡2x+sin2 x-48cos2 xsin⁡x

=cos-17cos⁡xcos⁡x+1-49cos2x1-cos2x

=cos-1cos⁡x-cos-17cos⁡x

=x-cos-17cos⁡x

Hence, the value of k=7

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

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