Question

The number of solutions of $2{\sin }^{2}x+{\sin }^{2}2x=2,xϵ[0,2\pi ]$, is

• A. 4
• B. 5
• C. 7
• D. 6

Answer 1

The correct option is D 6

Given equation is $2{\sin }^{2}x+{\sin }^{2}2x=2$

$2{\sin }^{2}x-1+{\sin }^{2}2x=1$

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

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

$\cos 2x+{\cos }^{2}2x=0$

$\cos 2x(\cos 2x+1)=0$

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

$2x=2n\pi \pm \frac{\pi }{2}$ or $2x=2n\pi \pm \pi$

So, there will be $6$ solutions from $[0,180]$