Question

The number of solution's of $2{\sin }^{2}x+{\sin }^{2}2x=2,x\in [0,2\pi ]$ are :

• A. $4$
• B. $5$
• C. $7$
• D. $6$

Answer 1

The correct option is D $6$

$(1-\cos 2x)+(1-{\cos }^{2}2x)=2$
$\Rightarrow \cos 2x(\cos 2x+1)=0$
$\Rightarrow \cos 2x=0$ or $\cos 2x=-1$
$\Rightarrow 2x=(2n+1)\frac{\pi }{2}$ or $2x=(2n-1)\pi ,n\in Z$
$\Rightarrow x=(2n+1)\frac{\pi }{4}\text{or}x=(2n-1)\frac{\pi }{2},n\in Z$
Hence, the solutions are $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4},\frac{\pi }{2},\frac{3\pi }{2}$