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

							x)2 = 3/42 cos2 2x + (2sinx = 3/4     24 os 2x + c2sinsin2 2x + (1 – sin2 2x)2 = 3/4     sin2 2x + 1 – 2 sin2 2x + sin4 2x  = 3/4or  sin4 2x – sin2 2x + 1/4 = 0    (sin2 2x – 1/2)2 = 0i.e. sin2 2x = 1/2  =  (1/)2   or   sin2 2x  =  sin2 /4  2x = n  /4  i.e.  x = n/2  /8Put n=0, then x = /8  [ x belongs to 0 to /2 or 90]Put n = 1, then x = /2  /8  =  5/8  or  3/8  [x = 5/8 isn’t possible for x belongs to 0 to /2]Hence /8 and 3/8 are the required solutions whosesum = /8 + 3/8 = /2 or 90 .

8 months ago
							  = 3/4x2 sin4 + x2 sin2sin2 2x + 1 – 2   = 3/4    2x + (1 – sin2 2x)2 2sin= ¾ 2x + (cos2 2x)2 2sinor  sin4 2x – sin2 2x + 1/4 = 0    (sin2 2x – 1/2)2 = 0i.e. sin2 2x = 1/2  =  (1/)2   or   sin2 2x  =  sin2 /4  2x = n  /4  i.e.  x = n/2  /8Put n=0, then x = /8  [ x belongs to 0 to /2 or 90]Put n = 1, then x = /2  /8  =  5/8  or  3/8  [x = 5/8 isn’t possible for x belongs to 0 to /2]Hence /8 and 3/8 are the required solutions whosesum = /8 + 3/8 = /2 or 90 .

8 months ago
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