How do you find the exact solutions of the equation #(sin2x+cos2x)^2=1# in the interval #[0,2pi)#?

Question

14207 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-25T14:37:32

Exact solution in interval #[0,2pi)# is #{0,pi/4,pi/2,(3pi)/4,pi,(5pi)/4,(3pi)/2,(7pi)/4}#

#(sin2x+cos2x)^2=1# can be expanded as

#sin^2(2x)+cos^2(2x)+2sin2xcos2x=1#

or #sin^2(2x)+cos^2(2x)+2sin2xcos2x=1#

or #1+sin4x=1#

i.e. #sin4x=0#

and hence #4x=npi# and #x=(npi)/4#

Hence, exact solution in interval #[0,2pi)# is #{0,pi/4,pi/2,(3pi)/4,pi,(5pi)/4,(3pi)/2,(7pi)/4}#