Tuesday, August 13, 2013

Solve: (sin(2x)+cos(2x))^2=1

After expanding and simplifying we can write 
%0D%0A%28sin2x%2Bcos2x%29%5E2%0D%0A 

as 

sin4x%2B1%0D%0A 

Therfore 
%0D%0A%28sin2x%2Bcos2x%29%5E2=1%0D%0A 

becomes 

sin4x%2B1=1%0D%0A 

and so 


sin4x=0%0D%0A 


which has solutions... 

x=%28n%2Api%29%2F4 

 where n is an integer. 

:)
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