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Question 66 of 71

Q.The number of real numbers in [0,2π][0, 2\pi] satisfying sin⁡4x−2sin⁡2x+1\sin^4x-2\sin^2x+1 is :

(a) 11
(b) 22
(c) ∞\infty
(d) 44
Puducherry TnboardTamil Nadu HSC (DGE) Board 2025MCQ· 1mImportance★★★★★
93% · 66/71 Questions
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The quartic in sin⁡x\sin x is a perfect square that forces sin⁡2x=1\sin^2x=1, which has exactly two solutions in [0,2π][0,2\pi].

  1. The equation is sin⁡4x−2sin⁡2x+1=0\sin^4x-2\sin^2x+1=0.
  2. Let t=sin⁡2xt=\sin^2x. Then t2−2t+1=0⇒(t−1)2=0⇒t=1t^2-2t+1=0\Rightarrow(t-1)^2=0\Rightarrow t=1.
  3. So sin⁡2x=1⇒sin⁡x=±1\sin^2x=1\Rightarrow\sin x=\pm1. …

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