Question

已知sin² 2x+sin2xcosx-cos2x=1, x∈(0,2/π）,求sinx,tanx

Answer 1

sin2x=2sinxcosx;cos2x+1=2(cosx)^2

=>(sin2x)^2+sin2xcosx=cos2x+1=2(cosx)^2

=> 2sinx(2sinx+1)(cosx)^2=2(cosx)^2

∵x∈(0,2/π）=>cosx≠0,且sinx>0

∴sinx(2sinx+1)=1

=>2(sinx)^2+sinx-1=0,解方程可得俩根-1和1/2

舍去-1,得sinx=1/2,且x=π/6

=>tanx=根号3

就这么多了,