已知2cosx+3sinx=(根号26)/2,求tan2x

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已知2cosx+3sinx=(根号26)/2,求tan2x

2cosx+3sinx=(√26)/2,
(2/√13)cosx+(3/√13)sinx=(√2)/2.
设2/√13=siny,3/√13=cosy,则
sinycosx+cosysinx=(√2)/2,sin(x+y)=(√2)/2,x+y=π/4+2kπ,x=π/4-y+2kπ,
tanx=tan(π/4-y)=[tan(π/4)-tany]/[1+tan(π/4)tany]=(1-2/3)/(1+2/3)=1/5.
∴tan2x=2tanx/[1-(tanx)^2]=(2/5)/[1-(1/5)^2]=5/12.