Question

Solve $2{\sin }^{2}x+{\sin }^{2}2x=2$

• A. $x=2n\pi +\frac{\pi }{2}$ where $n\in Z$
• B. $x=m\pi \pm \frac{\pi }{8}$, $mϵZ$, where $m,\in Z$
• C. $x=m\pi \pm \frac{\pi }{4}$, $mϵZ$, where $m,\in Z$
• D. $x=2n\pi +\frac{\pi }{4}$ where $n\in Z$

Answer 1

Answer: (A), (C)

The correct options are
A $x=2n\pi +\frac{\pi }{2}$ where $n\in Z$
C $x=m\pi \pm \frac{\pi }{4}$, $mϵZ$, where $m,\in Z$

We have $2{\sin }^{2}x+{\sin }^{2}2x=2$

$2{\sin }^{2}x+{\left(2\sin x\cos x\right)}^2=2$

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

$2{\sin }^{2}x{\cos }^{2}x-(1-{\sin }^{2}x)=0$

$2{\sin }^{2}x{\cos }^{2}x-{\cos }^{2}x=0$

${\cos }^{2}x(2{\sin }^{2}x-1)=0$

${\cos }^{2}x=0$ or ${\sin }^{2}x=\frac{1}{2}$

$\Rightarrow$ $x=2n\pi +\frac{\pi }{2}$

${\sin }^{2}x={\sin }^2\frac{\pi }{4}$

$=2n\pi +\frac{\pi }{2}$ or $x=m\pi \pm \frac{\pi }{4}$, $m\in Z$, where $m,n\in Z$