2 sinx cosx+1-2sin2x=1 2 sinx cosx-2sin2x=0 2 sinx (cosx-sinx)=0 And we have two opportunities: either sinx=0 or cosx-sinx=0 1) sinx=0 x=pi n, n -integer 2) cosx-sinx=0 tanx=1 x=pi/4+2pi k, k-integer So the answer is: pi n, pi/4+2pi k, where n,k-integers.
Finding a professional expert in "partial differential equations" in the advanced level is difficult.
You can find this expert in "Assignmentexpert.com" with confidence.
Exceptional experts! I appreciate your help. God bless you!
Comments