(1)2/3sin2x+1/4cos2x
=[2/3*2sinxcosx+1/4(cos²x-sin²x)]/(sin²x+cos²x) 两边同除以cos²x
=(4/3tanx+1/4-1/4tan²x)/(tan²x+1)
∵tanx=2
∴原式=（4/3*2+1/4-1/4*2²）/（2²+1）=23/60
(2)2sin2x-sinxcosx+cos2x
=2*2sinxcosx-sinxcosx+cos²x-sin²x
=（3sinxcosx+cos²x-sin²x）/(sin²x+cos²x) 两边同除以cos²x
=（3tanx+1-tan²x)/(tan²x+1)
∵tanx=2
∴原式=(3*2+1-2²)/（2²+1）=3/5