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.
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