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Grade: 12
        
the sum of all values of x belongs from 0 to 90 dedree satisfying sin2 2x+cos4 2x=3/4
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
4 months ago

Answers : (2)

Samyak Jain
327 Points
							x)2 = 3/42 cos2 2x + (2sinx = 3/4   \Rightarrow  2os 2x + c2sin
sin2 2x + (1 – sin2 2x)2 = 3/4   \Rightarrow  sin2 2x + 1 – 2 sin2 2x + sin4 2x  = 3/4
or  sin4 2x – sin2 2x + 1/4 = 0  \Rightarrow  (sin2 2x – 1/2)2 = 0
i.e. sin2 2x = 1/2  =  (1/\small \sqrt{2})2   or   sin2 2x  =  sin2 \pi/4
\therefore  2x = n\pi \pm \pi/4  i.e.  x = n\pi/2 \pm \pi/8
Put n=0, then x = \pi/8  [\because x belongs to 0 to \pi/2 or 90\degree]
Put n = 1, then x = \pi/2 \pm \pi/8  =  5\pi/8  or  3\pi/8  [x = 5\pi/8 isn’t possible for x belongs to 0 to \pi/2]
Hence \pi/8 and 3\pi/8 are the required solutions whose
sum = \pi/8 + 3\pi/8 = \pi/2 or 90\degree .
4 months ago
Samyak Jain
327 Points
							  = 3/4x2 sin4 + x2 sin2sin2 2x + 1 – 2   \Rightarrow= 3/4    2x + (1 – sin2 2x)2 2sin= ¾ 2x + (cos2 2x)2 2sin
or  sin4 2x – sin2 2x + 1/4 = 0  \Rightarrow  (sin2 2x – 1/2)2 = 0
i.e. sin2 2x = 1/2  =  (1/\small \sqrt{2})2   or   sin2 2x  =  sin2 \pi/4
\therefore  2x = n\pi \pm \pi/4  i.e.  x = n\pi/2 \pm \pi/8
Put n=0, then x = \pi/8  [\because x belongs to 0 to \pi/2 or 90\degree]
Put n = 1, then x = \pi/2 \pm \pi/8  =  5\pi/8  or  3\pi/8  [x = 5\pi/8 isn’t possible for x belongs to 0 to \pi/2]
Hence \pi/8 and 3\pi/8 are the required solutions whose
sum = \pi/8 + 3\pi/8 = \pi/2 or 90\degree .
4 months ago
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