Q25 / 71 · NTA Abhyas (JEE Main) 2020 · Mathematics · Inverse Trigonometric Functions

The value(s) of x satisfying the equation sin-1⁡1-x-2sin-1⁡x=π2 is/are

  • Correct answer
    A
    0
  • B
    12
  • C
    0,12
  • D
    -12

Solution

We have,
sin-1⁡1-x-2sin-1⁡x=π2
⇒sin-1⁡1-x=π2+2sin-1⁡x
⇒1-x=sin⁡π2+2sin-1⁡x
⇒1-x=cos⁡2sin-1⁡x
⇒1-x=cos⁡cos-1⁡1-2x2∵2sin-1⁡x=cos-1⁡1-2x2
⇒1-x=1-2x2
⇒x=2x2⇒x2x-1=0⇒x=0,12
For, x=12, we have
LHS =sin-1⁡1-x-2sin-1⁡x
=sin-1⁡12-2sin-1⁡12=-sin-1⁡12=-π6≠R.H.S.
So, x=12 is not a root of the given equation.
Clearly, x=0 satisfies the equation
Here, x=0 is the root of the given equation.

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

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