Skip to content
Question 60 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 2023MCQ· 1mImportance★★★★★
85% · 60/71 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Recognising sin⁡4x−2sin⁡2x+1\sin^4x-2\sin^2x+1 as a perfect square in sin⁡2x\sin^2x reduces the equation to cos⁡x=0\cos x=0, which has 2 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. This factors as a perfect square: sin⁡4x−2sin⁡2x+1=(sin⁡2x−1)2\sin^4x-2\sin^2x+1=(\sin^2x-1)^2.
  3. So (sin⁡2x−1)2=0⇒sin⁡2x=1⇒cos⁡2x=1−sin⁡2x=0⇒cos⁡x=0(\sin^2x-1)^2=0\Rightarrow\sin^2x=1\Rightarrow\cos^2x=1-\sin^2x=0\Rightarrow\cos x=0. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.