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MHT CET202527 Apr 2025Evening ShiftMathematicsTrigonometric EquationsActual

If 1- = 2 , then the value of is

Options

  1. A2 n , 4 n , n Z
  2. Bn 2 , n 3 , n Z
  3. C(2 n+1) 2 , n Z
  4. D(2 n-1) 4 , n Z

Correct answer

A. 2 n , 4 n , n Z

Step-by-step solution

Beginning with 1- = 2 , apply the half-angle identity 1- = 2 ^2 2 and the double-angle identity = 2 2 2 to rewrite the equation: 2 ^2 2 = 2 ^2 2 2 Subtract the right side and factor: 2 ^2 2 (1- 2 ) = 0 This implies either 2 = 0 or 2 = 1 . The solution 2 = 0 yields 2 = n or = 2n for integer n . The solution 2 = 1 yields 2 = 2k or = 4k for integer k . Since all values of the form 4k are already contained in the set 2n , the general solution is = 2n for integer n . Comparing with the options, only option A includes th

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