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Question

Number of solution(s) of equation sin3xcosx+sin2xcos2x+sinxcos3x=1  x[0,2π] is :
  1. 4
  2. 2 
  3. 1 
  4. 0 


Solution

The correct option is D 0 
sin3xcosx+sin2xcos2x+sinxcos3x=1
sinxcosx[sin2x+sinxcosx+cos2x]=112sin2x[1+12sin2x]=1
Let sin2x=t
t(2+t)=4t2+2t4=0t=2±252=1±5sin2x=15<1  (not possible)
or sin2x=51>1  (not possible)
Hence no solution
 

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