Question

The general solution(s) of ${\cos }^{2}2x-{\sin }^{2}x=0$ can be

• A. $(2n+1)\frac{\pi }{6},n\in Z$
  
• B. $(2n+1)\frac{\pi }{4},n\in Z$
  
• C. $(2n+1)\frac{\pi }{2},n\in Z$
• D. $(2n+1)\pi ,n\in Z$

Answer 1

Answer: (A), (C)

The correct options are
A $(2n+1)\frac{\pi }{6},n\in Z$

C $(2n+1)\frac{\pi }{2},n\in Z$

${\cos }^{2}2x-{\sin }^{2}x=0$
$\Rightarrow \cos 3x\cos x=0$
$\Rightarrow \cos 3x=0\text{or}\cos x=0$
$\Rightarrow 3x=(2n+1)\frac{\pi }{2}\text{or}x=(2n+1)\frac{\pi }{2},n\in Z$
$\Rightarrow x=(2n+1)\frac{\pi }{6}\text{or}x=(2n+1)\frac{\pi }{2},n\in Z$