Question

${cos}^{2}2x+2{cos}^{2}x=1,xϵ(-\pi ,\pi )$, then $x$ can take the values

• A. $\pm \frac{\pi }{2}$
• B. $\pm \frac{\pi }{4}$
• C. $\pm \frac{3\pi }{8}$
• D. Non of these

Answer 1

The correct option is B $\pm \frac{\pi }{4}$

Given $x\in (-\pi ,\pi )⟹2x\in (-2\pi ,2\pi )$

${\cos }^{2}2x+2{\cos }^{2}x=1$

$⟹{\cos }^{2}2x+2{\cos }^{2}x-1=0$

$⟹{\cos }^{2}2x+(2{\cos }^{2}x-1)=0$

$⟹{\cos }^{2}2x+\cos 2x=0$

$⟹\cos 2x(\cos 2x+1)=0$

So $\cos 2x=0$ or $-1$

$⟹2x=\pm \frac{\pi }{2},\pm \frac{3\pi }{2}$ or $\pm \pi$

$⟹x=\pm \frac{\pi }{4},\pm \frac{3\pi }{4}$ or $\pm \frac{\pi }{2}$