Question

Solve the equation for general solution $2{\sin }^{2}x+{\sin }^{2}2x=2$?

• A. $n\pi +\frac{\pi }{2},(2m+1)\frac{\pi }{2},n\pi -\frac{\pi }{2}$
  
• B. $n\pi -\frac{\pi }{4},(2m+1)\frac{\pi }{2},n\pi -\frac{\pi }{8}$
• C. $n\pi +\frac{\pi }{4},(2m+1)\frac{\pi }{2},n\pi -\frac{\pi }{4}$
  
• D. $n\pi +\frac{\pi }{4},(m+1)\frac{\pi }{2},n\pi -\frac{\pi }{2}$
  

Answer 1

Answer: (C)

$n\pi +\frac{\pi }{4},(2m+1)\frac{\pi }{2},n\pi -\frac{\pi }{4}$

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

$2{\sec }^{2}x+(2\sec x\cos x{)}^2=2$

$2{\sec }^{2}x+4{\sec }^{2}x{\cos }^{2}x=2$

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

${\sin }^{2}x+2{\sin }^{2}x-2{\sin }^{4}x=1$

$2{\sin }^{4}x-3{\sin }^{2}x+1=0$

$2{\sin }^{4}x-2{\sin }^{2}x-{\sin }^{2}x+1=0$

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

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

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

$n\pi +(-1{)}^n\frac{\pi }{4};(2m+1)\frac{\pi }{2}$

$n\pi +\frac{\pi }{4},n\pi -\frac{\pi }{4};(2m+1)\frac{\pi }{2}$