Question

give the most general solutions to equations. . 2sinxcosx-sin(2x)cos(2x)=0. . A. simplify the first expression using double angle identity for sine.. B. factor the left side of equatioin. C. solve the factored equation

Answer 1

2sinxcosx - sin(2x)cos(2x) = 0 

Part I 
The double angle identity for sine states that sin(2x) = 2sinxcosx 
Thus we get: 
sin(2x) - sin(2x)cos(2x) = 0 

Part II 
sin(2x)(1 - cos(2x)) = 0 

Part III 
Either sin(2x) = 0 or 
1 - cos(2x) = 0 
=> cos(2x) = 1 

For sin(2x) = 0, this is true for 
2x = n(pi) where n = 0, 1, 2, .... 
x = n(pi/2) 

For cos(2x) = 1, this is true for 
2x = n(pi) where n = 0, 2, 4, .... 
x = n(pi/2)


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